Bourbaki

Exercise 3, § 2

Integration, V, p. 95

Let T be the compact interval [0,1][0, 1] of R\mathbf{R}; for every integer n>0n > 0, let λn\lambda_n be the measure

f↦1n∫01f(x)sin⁡2nπx dx f \mapsto \frac{1}{n} \int_0^1 f(x) \sin 2n \pi x \, dx
on T. Show that, in the space M(T)\mathscr{M}(T) equipped with the vague topology (or with the quasi-strong topology (Ch. III, §1, Exer. 8)), the sequence of measures (λn)(\lambda_n) is summable, but the sequence (∣λn∣)(|\lambda_n|) is not. (Observe that if αn=n⋅λn(f)\alpha_n = n \cdot \lambda_n(f) then ∑nαn2<+∞\sum_n \alpha_n^2 < +\infty, by making use of Bessel’s inequality, and deduce from this that ∑n∣λn(f)∣<+∞\sum_n |\lambda_n(f)| < +\infty by the Cauchy–Schwarz inequality.)

Solution

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