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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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Bounded linear maps commute with Bochner integration

Statement

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

T(∫Ef dμ)=∫ETf dμ.

Facts & Assumptions

[L1]

A bounded linear operator satisfies ∥Tx∥≤C∥x∥ 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.

1.1givenL2choose

Restrict a defining approximation to the measurable set. Choose integrable simple sn with ∫∥f−sn∥→0. Then 1ETsn is an integrable Y-valued simple function: every nonzero level is a finite union of level sets of 1Esn.

2.1L1L2step 1.1

Prove Bochner integrability after applying T. By [L1], ∫∥1ETf−1ETsn∥≤C∫E∥f−sn∥→0. Thus [L2] makes 1ETf Bochner integrable.

3.1L1L2L3step 1.1step 2.1∎

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.

Depends on

Used by

…and 10 more results.

Dependency tree · two levels

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Sources