2002
DOI: 10.1090/s0002-9947-02-03203-8
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Lebesgue type decomposition of subspaces of Fourier-Stieltjes algebras

Abstract: Abstract. Let G be a locally compact group and let A(G) and B(G) be the Fourier algebra and the Fourier-Stieltjes algebra of G, respectively. For any unitary representation π of G, let Bπ(G) denote the w * -closed linear subspace of B(G) generated by all coefficient functions of π, and B 0 π (G) the closure of Bπ (G) ∩ Ac(G), where Ac(G) consists of all functions in A(G) with compact support. In this paper we present descriptions of B 0 π (G) and its orthogonal complement B s π (G) in Bπ(G), generalizing a rec… Show more

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Cited by 9 publications
(4 citation statements)
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“…As already remarked, eA * * is isometrically isomorphic to the multiplier algebra M (A). The Lebesgue decomposition theorem and its extension to the Fourier-Stieltjes algebra, see [14] and [13], show that group algebras on compact groups and Fourier algebras on discrete amenable groups are both standard wscai algebras. Theorem 1.9.…”
Section: Introductionmentioning
confidence: 99%
“…As already remarked, eA * * is isometrically isomorphic to the multiplier algebra M (A). The Lebesgue decomposition theorem and its extension to the Fourier-Stieltjes algebra, see [14] and [13], show that group algebras on compact groups and Fourier algebras on discrete amenable groups are both standard wscai algebras. Theorem 1.9.…”
Section: Introductionmentioning
confidence: 99%
“…Our assumption implies that there is a continuous unitary representation σ of G that is disjoint from λ G such that π = λ G ⊕σ and A = A(G)⊕ 1 A s (G), where A s (G) = A σ is what we shall call the singular part of A [1, Proposition 3.12, Corollary 3.13 and Theorem 3.18]. When A = B(G), this is the Lebesgue decomposition of B(G) that was studied in [25] and [22]. With the following lemmas, we record more useful facts for future reference.…”
Section: Preliminary Resultsmentioning
confidence: 99%
“…We begin with a few preliminaries, starting with the following Lemma that basically follows from [22,Lemma 1.1]. Recall (see [30] or [42]) that B(Γ) = A(Γ) ⊕ B s (Γ) where B s (Γ) is a closed translation invariant subspace of B(Γ). If φ = φ a + φ s ∈ B(Γ), then φ = φ a + φ s .…”
Section: Is Reflexive If and Only If It Has Finite Dimensionmentioning
confidence: 99%