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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27
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Strong continuity of left and modular right translations on L1 and L2

Statement

Assume AC. Let G be an LCH group with a fixed left Haar measure μ, ΔG its modular function, and for g∈G define Lgf(x):=f(g−1x),Rgf(x):=ΔG(g)1/2f(xg). Then g↦Lg is strongly continuous on L1(G) and on L2(G), and g↦Rg is strongly continuous on L2(G): for every f in the space and every ϵ>0 there is a neighbourhood V of e in G with ∥Lgf−f∥p<ϵ, respectively ∥Rgf−f∥2<ϵ, for every g∈V — and then, for the general point g0, ∥Lgf−Lg0f∥p→0 and ∥Rgf−Rg0f∥2→0 as g→g0, neighbourhoods rather than sequences being used throughout since G need not be first countable.

Facts & Assumptions

Given: An LCH group G with a fixed left Haar measure μ, its modular function ΔG, the complex spaces L1(G),L2(G) with norms ∥⋅∥p, and AC.

[F1]

For p∈{1,2} the complex space Lp(G)=Lp(G,μ;C) consists of the a.e. classes of measurable complex functions with ∥f∥p=(∫G∣f∣p dμ)1/p<∞, and Cc(G)=Cc(G;C) (Complex Haar L^p spaces and compactly supported functions).

[F2]

Left invariance: μ(aE)=μ(E) for every Borel E and a∈G, so ∫GH(ax) dμ(x)=∫GH(x) dμ(x) for nonnegative Borel H (Left Haar integral and left Haar measure); and for each g there is c(g)>0 with ∫GF(xg) dμ(x)=c(g)∫GF dμ(x) for every nonnegative Borel F, the scalar being the unique one with that property and equal to ΔG(g−1) (Right translation scales left Haar measure, Uniqueness of left Haar measure up to scale, Modular function of a locally compact group).

[F3]

ΔG is a continuous homomorphism G→R>0 with ΔG(e)=1, so g↦ΔG(g)1/2 is continuous at e with value 1, and ΔG(g)ΔG(g−1)=1 (The modular function is a continuous homomorphism).

[F4]

For f∈Cc(G) the maps a↦Laf and a↦ρaf, where ρaf(x):=f(xa) is the unscaled right translate, are continuous in uniform norm at every a0∈G, with all supports contained in one fixed compact set on a neighbourhood of a0; both translates lie in Cc(G) (Translations preserve compactly supported continuous functions). The modular right translate of the Statement is Raf=ΔG(a)1/2ρaf, whose scalar factor is continuous by [F3].

[F5]

Cc(G) is dense in L1(G) and in L2(G) (Completeness of the complex Haar L1 and L2 spaces and density of Cc).

[F6]

A left Haar measure is finite on compact sets (Left Haar integral and left Haar measure).

[A1]

AC is assumed in the choice-function form of the cited definition; it is inherited here through the density statement [F5] (The Axiom of Choice).

Proof

technique · direct
1.1

Isometries. Let f∈L1(G). By [F2] both ∫G∣f(g−1x)∣ dμ(x)=∫G∣f(y)∣ dμ(y) and, since ∣Rgf(x)∣=ΔG(g)1/2∣f(xg)∣, ∥Rgf∥1=ΔG(g)−1/2∥f∥1. For f∈L2(G) the same substitution gives ∥Lgf∥2=∥f∥2, while ∥Rgf∥22=∫GΔG(g)∣f(xg)∣2 dμ(x)=ΔG(g)c(g)∥f∥22=∥f∥22 by [F2] and ΔG(g)ΔG(g−1)=1 from [F3]. So Lg is isometric on L1(G) and on L2(G), and Rg is isometric on L2(G) (it scales the L1 norm by ΔG(g)−1/2).

F1F2F3
1.2

Uniform-norm continuity at e for compactly supported functions. Let f∈Cc(G). By [F4] applied at a0=e there are a neighbourhood V0 of e and a compact C⊆G containing the supports of Laf, ρaf and f for every a∈V0, with ∥Laf−f∥∞→0 and ∥ρaf−f∥∞→0 as a→e. The scalar ΔG(a)1/2 tends to 1 by [F3], and ∥Raf−f∥∞≤ΔG(a)1/2∥ρaf−f∥∞+∣ΔG(a)1/2−1∣ ∥f∥∞⟶0. Since this scalar is positive, supp⁡Raf=supp⁡ρaf⊆C. Thus for every ϵ>0 some neighbourhood V⊆V0 makes both ∥Laf−f∥∞ and ∥Raf−f∥∞ less than ϵ for all a∈V.

F3F4
2.1

Compactly supported convergence in Lp. With f, C and V as in step 1.2, for a∈V the difference Laf−f is supported in the compact set C, so ∥Laf−f∥1≤∥Laf−f∥∞ μ(C) and ∥Laf−f∥2≤∥Laf−f∥∞ μ(C)1/2, both tending to 0 as a→e by the uniform bound of step 1.2, with μ(C)<∞ by [F6]; the same estimates hold for Raf−f. Thus La→id strongly on Cc(G) for p=1,2 and Ra→id strongly on Cc(G) for p=2.

F6step 1.2
3.1

Left translations on Lp(G). Fix f∈Lp(G) with p∈{1,2} and ϵ>0. By [F5] under the AC of [A1] choose h∈Cc(G) with ∥f−h∥p<ϵ/3, and by step 2.1 choose a neighbourhood V of e with ∥Lah−h∥p<ϵ/3 for a∈V. For a∈V the isometry of step 1.1 gives ∥Laf−f∥p≤∥La(f−h)∥p+∥Lah−h∥p+∥h−f∥p<3⋅ϵ/3=ϵ, so La→id strongly at e on Lp(G). For a general g0∈G, the definition Laf(x)=f(a−1x) gives LaLb=Lab, hence Lg=Lg0Lg0−1g. Therefore ∥Lgf−Lg0f∥p=∥Lg0(Lg0−1gf−f)∥p=∥Lg0−1gf−f∥p, by the isometry of step 1.1. This is <ϵ once g0−1g∈V, i.e. for g in the neighbourhood g0V of g0.

A1F5step 1.1step 2.1
3.2

Right translations on all of L2(G). Fix f∈L2(G) and ϵ>0, choose h∈Cc(G) with ∥f−h∥2<ϵ/3 and a neighbourhood V of e with ∥Rah−h∥2<ϵ/3 for a∈V, by step 2.1 and [F5]. Since Ra is isometric on L2(G) by step 1.1, ∥Raf−f∥2≤∥Ra(f−h)∥2+∥Rah−h∥2+∥h−f∥2<ϵ for a∈V. For a general g0, Rg=Rg0Rg0−1g because ΔG(g0)1/2ΔG(g0−1g)1/2=ΔG(g)1/2 by multiplicativity [F3], so ∥Rgf−Rg0f∥2=∥Rg0(Rg0−1gf−f)∥2=∥Rg0−1gf−f∥2<ϵ for g∈g0V, again by the isometry of step 1.1.

F3F5step 1.1step 2.1
4.1

Collecting the statements: step 3.1 gives strong continuity of g↦Lg on L1(G) and on L2(G) at every point, and step 3.2 gives strong continuity of g↦Rg on L2(G) at every point. ∎

step 3.1step 3.2

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