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DefinitionDefinition: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)audited 2026-10-08
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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.

Reiter's condition (P1)

Definition

Let G be a locally compact Hausdorff group with fixed left Haar measure μ, and put P:={f∈L1(G):f≥0, ∥f∥1=1}, where f≥0 means that the class has a real-valued representative that is nonnegative almost everywhere. For g∈G define (Lgf)(x):=f(g−1x) on almost-everywhere classes. For compact Q⊆G and f∈P, set ΔQ(f):=sup⁡({0}∪{∥Lgf−f∥1:g∈Q}). Then 0≤ΔQ(f)≤2, and Δ∅(f)=0. The group G satisfies Reiter's condition (P1) if for every compact Q⊆G and every ε>0 there is f∈P with ΔQ(f)≤ε.

Equivalently, there is a net (fi)i∈I in P with its inherited L1-norm topology such that for every compact Q and every ε>0 some i0∈I satisfies ΔQ(fi)≤ε for all i⪰i0; this is uniform convergence to zero on compact subsets. The set P is convex and LgP=P for every g∈G. Only left translates are used, the measure μ is fixed, and no compactness assumption on G is made.

Facts & Assumptions

Given: A locally compact Hausdorff group G with a fixed left Haar measure μ.

[A1]

For each g∈G, the map Tg(x)=g−1x is Borel measurable and measure-preserving: Tg−1(E)=gE and μ(gE)=μ(E) for Borel E⊆G. Thus it induces Lgf=f∘Tg on almost-everywhere classes (Left Haar integral and left Haar measure, A continuous map has Borel preimages of Borel sets, Measure-preserving transformations and systems).

[F1]

L1(G) consists of complex measurable almost-everywhere classes with ∥f∥1=∫G∣f∣ dμ; if f≥0, then ∫Gf=∥f∥1 (Complex Haar L^p spaces and compactly supported functions).

[F2]

Integrals are invariant under measure-preserving maps, and the integral is complex-linear on L1(G) (Integral invariance under measure-preserving maps, The Lebesgue integral is linear on L1(μ)).

[F3]

A net is a function indexed by a nonempty directed preorder; antisymmetry is not required (Directed preorders and nets).

Proof

technique · direct
1.1A1F1F2algebra

If f∈P, then [A1] makes Lgf well-defined on classes and preserves nonnegativity. By [F2], ∥Lgf∥1=∫G∣f∘Tg∣ dμ=∫G∣f∣ dμ=1. Thus LgP⊆P; applying the same argument to g−1 and using Lg−1Lg=I gives equality. Also, for every g∈G, ∥Lgf−f∥1≤∥Lgf∥1+∥f∥1=2, so the supremum defining ΔQ(f) is finite and lies in [0,2], including the empty-test value zero.

1.2F1F2constructalgebra

For f,h∈P and t∈[0,1], choose nonnegative real representatives. Their convex combination is nonnegative and, by [F2], ∥tf+(1−t)h∥1=∫G(tf+(1−t)h) dμ=t∥f∥1+(1−t)∥h∥1=1. Therefore tf+(1−t)h∈P and P is convex.

2.1F3step 1.1given

If (fi)i∈I is a net satisfying the compact-uniform condition, then for any compact Q and ε>0 its defining eventual estimate supplies i0 with ΔQ(fi)≤ε for all i⪰i0. In particular fi0∈P is a witness to Reiter's condition.

3.1F3constructalgebra∎

Conversely, assume Reiter's condition. Let I be the set of all triples (Q,ε,f) with Q compact, ε>0, f∈P, and ΔQ(f)≤ε. Order them by (Q,ε,f)⪯(Q′,ε′,f′) exactly when Q⊆Q′ and ε′≤ε. It is nonempty, since the condition at the compact singleton {e} and ε=1 supplies a witness. It is directed: for two indices apply the condition to the compact union of their test sets, which is compact as a finite union, and the positive minimum of their tolerances, obtaining a witness that gives a common upper bound. By [F3], the third-coordinate map i↦fi is a net. Given any compact Q and ε>0, the condition supplies i0=(Q,ε,f0); every i⪰i0 then has ΔQ(fi)≤ΔQi(fi)≤εi≤ε. Every index carries its own witness, so no global choice function is used.

Depends on

Used by

Dependency tree · two levels

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Sources