Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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The dual of ell-infinity is ba

Statement

The map

Φ:()ba(P(N)),Φ(φ)(A):=φ(1A),

is a linear isometric isomorphism. Its inverse sends ν to the finitely additive integral Iν.

Facts & Assumptions

[L1]

For each finite-variation charge, Iν is a bounded functional and Iν=ν(N) (The finitely additive integral is well-defined and isometric).

[L2]

The dual consists of bounded scalar-valued linear functionals with the operator norm (The dual space X^* of a normed space and its dual norm).

Proof

technique · direct

Given: The objects and hypotheses in the Statement.

1.1

Let φ() and put νφ(A)=φ(1A). Linearity and 1AB=1A+1B give finite additivity. For a finite partition (Aj) choose scalar phases cj with cjνφ(Aj)=νφ(Aj). Then jcj1Aj=1 and [L2] gives

givenL2

jνφ(Aj)=φ(jcj1Aj)φ.

Thus νφba and νφbaφ. [L2, finite additivity, phases]

2.1

By construction, Iνφ(1A)=φ(1A). [given, L1, step 1.1] Linearity gives equality on finite-range sequences, and density plus boundedness gives Iνφ=φ on .

L1step 1.1algebra
3.1

Conversely, Φ(Iν)(A)=Iν(1A)=ν(A), so the two maps are [given, L1, step 2.1] inverse. Finally [L1] gives Iν=νba, proving isometry and completing both surjectivity and injectivity.

L1step 2.1

Depends on

Used by

Dependency tree · two levels

9 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