Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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.

Bounded linear maps commute with Bochner integration

Statement

Let X,Y be Banach spaces, let T:XY be bounded and linear, and let f be Bochner integrable. Then Tf is Bochner integrable and, for every measurable E,

T(Efdμ)=ETfdμ.

Facts & Assumptions

[L1]

A bounded linear operator satisfies TxCx for some finite C (A bounded linear operator between normed spaces).

[L2]

A Bochner integral is the norm limit of integrals of an L1-approximating simple sequence (Bochner-integrable function).

[L3]

The Banach-valued simple integral is linear and representation-independent (The Banach-valued simple integral is well defined).

Proof

technique · direct

Given: T,f,E as in the Statement.

1.1

Restrict a defining approximation to the measurable set. Choose integrable simple sn with fsn0. Then 1ETsn is an integrable Y-valued simple function: every nonzero level is a finite union of level sets of 1Esn.

givenL2choose
2.1

Prove Bochner integrability after applying T. By [L1], 1ETf1ETsnCEfsn0. Thus [L2] makes 1ETf Bochner integrable.

L1L2step 1.1
3.1

Commute T with the defining limit. [L1, L2, L3, step 1.1, step 2.1] For each simple sn, finite linearity in [L3] gives T(Esn)=ETsn. Boundedness makes T norm-continuous, so taking limits in this equality and using [L2] proves the displayed identity. If T=0, f=0, or E=, both sides are explicitly zero.

L1L2L3step 1.1step 2.1

Depends on

Used by

Dependency tree · two levels

10 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