By Scott R.F.

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Let μ be a probability measure on T. The following assertions are equivalent : a) b) c) d) μ is supported by a weak-Dirichlet set. lim supγ →∞ |μˆ (γ )| = 1. 1 is the limit in L1 (μ ) of a sequence (γ j ) ⊂ Γ which tends to infinity in Γ . There exists an idempotent h ∈ Γ , h = 1, such that h μ = μ . Proof. Let h ∈ Γ , h = 1, such that h μ = μ . In particular, we have 1 = μˆ (h) ≤ lim sup |μˆ (γ )| ≤ 1 γ →∞ and d) implies b). Now suppose that b) holds. Then, there exists a sequence (γ j ) ⊂ Γ which tends to infinity in Γ and satisfies lim sup |μˆ (γ j )| = 1.

Under assumption (b) we have to prove that U has simple spectrum. If this is not the case, for any h ∈ H, there would exist g ∈ H such that [U, g] ⊥ [U, h], so that, by (b), σh ⊥ σg . But this would contradict the following result ensuring the existence of maximal elements : There exists h ∈ H such that σg σh for every g ∈ H. 4. U has simple Lebesgue spectrum on H if H = [U, h] for some h ∈ H with σh ∼ m. 10. U has simple Lebesgue spectrum on H if and only if there exists h ∈ H such that (U n h)n∈Z forms an orthonormal basis of H.

Proof. 7) and thanks to the bilinearity property of σ f ,g , we check that σφ (U)( f +g)−φ (U) f −φ (U)g = |φ |2 (σ f +g + σ f + σg − σ f +g, f − σ f , f +g − σ f +g,g − σg, f +g + σ f ,g + σg, f ) = 0 so that ||φ (U)( f + g) − φ (U) f − φ (U)g||2H = ||σφ (U)( f +g)−φ (U) f −φ (U)g || = 0 In the same way, ||φ (U)(λ f ) − λ φ (U) f ||H = ||σφ (U)(λ f )−λ φ (U) f || = 0, and φ (U) is linear. Finally, ||φ (U) f ||2H = ||σφ (U) f || = |||φ |2 σ f || ≤ ||φ ||2∞ || f ||2H gives the claimed bound. 2 (Spectral representation theorem).