Bourbaki

Exercise 64, § 1

Tag 02M4, permanent. Cite it as [Bourbaki, Tag 02M4].

Spectral Theories, II, p. 300

Let μ\mu be a measure on T\mathbf{T}.

a) For every x∈Tx\in \mathbf{T}, one has

1 2iπhx2i\pi hx

μ({x})=\mu(\{x\}) = lim ∑μ(h)e\sum\mu(h)e.

H→+∞2H+1∣h∣⩽H _{H\rightarrow+\infty}2H + 1_{|h|\leqslant H}

b) Let us write μ=μd+μa\mu=\mu_d+\mu_a, where μa\mu_a is an atomic measure and μd\mu_d a diffuse measure (INT, V, §5, no10^o10, Proposition 15). Let S be the support of μa\mu_a. One has

limH→+∞12H+1∑∣μ^(h)∣2=∑∣μ({s})∣2_{H\rightarrow+\infty}\frac{1}{2H + 1}\sum|\widehat{\mu}(h)|^2=\sum|\mu(\{s\})|^2

∣h∣⩽Hx∈S|h|\leqslant Hx\in S

(consider (μ∗μ~)({0})(\mu*\widetilde{\mu})(\{0\})).

c) Suppose that μ\mu is a probability measure on T\mathbf{T}. The following conditions are equivalent:

(i) There exists x∈Tx\in \mathbf{T} such that μ=εx\mu=\varepsilon_x;

(ii) One has lim∣h∣→+∞∣μ^(h)∣2=1_{|h|\rightarrow+\infty}|\widehat{\mu}(h)|^2= 1 ;

(iii) One has

limH→+∞12H+1∑∣μ^(h)∣2=1_{H\rightarrow+\infty}\frac{1}{2H + 1}\sum|\widehat{\mu}(h)|^2= 1.

∣h∣⩽H|h|\leqslant H
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Machine translation of the English, by gpt-5-6. Not checked by a person.