Haar measure and continuous representations of locally compact abelian groups

Abstract

Let L(X) be the algebra of all bounded operators on a Banach space X, and let θ : G → L(X) be a strongly continuous representation of a locally compact and second countable abelian group G on X. Set σ 1 (θ(g)) := {λ/|λ| | λ ∈ σ(θ(g))}, where σ(θ(g)) is the spectrum of θ(g), and let Σ θ be the set of all g ∈ G such that σ 1 (θ(g)) does not contain any regular polygon of T (by a regular polygon we mean the image under a rotation of a closed subgroup of the unit circle T different from {1}). We prove that θ is uniformly continuous if and only if Σ θ is a non-null set for the Haar measure on G.

Representations of Polish groups and continuity

“…One can see ϕ is a discontinuous morphism from G to T and (ϕ(g)) 3 = 1 for g in G. So, if V is an open subset in T containing no cubic root of unity, ϕ −1 (V ) is empty. Nevertheless, if we restrict the classes of the open subsets of T and subsets of G considered (in a manner which depends on the morphism ϕ or, more precisely, on Γ ϕ ), we can obtain a satisfactory generalization of the result, valid when G = R. Such theorems are proved in [3] and [28]. The approach below unifies the previous results and provides some details and generalizations.…”

“…We use the theorem obtained in this way to prove automatic continuity of Baire morphisms from a Polish group to a locally compact group. In the final part, the aim is to generalize and unify the description of spectra of elements in the range of a strongly continuous but not norm continuous representation, given in [3], [24] and [28], thus showing interesting applications of the continuity criteria provided in Section 2.…”

“…We first assume that X is separable. If θ is not norm continuous, by [28,Proposition 2.2] in case (1), and by Theorem 2.7 if G is not locally compact but θ is norm bounded (with A a commutative subalgebra of L(X) containing θ(G)), there is a χ ∈ Â such that χ • θ and clearly also χ 1 • θ is discontinuous, and so Γ χ 1 •θ = {1}. By the preceding lemma, we have…”