Bourbaki

Exercise 35, § 1

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

Spectral Theories, II, p. 277

a) For every integer N⩾1N\geqslant 1, every family (an)1⩽n⩽N(a_n)_{1\leqslant n\leqslant N} of complex numbers and every integer H⩾1H\geqslant 1, one has

∣∑Nan∣2⩽(1+N+1H)∑(1−∣h∣H)∑an+ha‾n |\sum^Na_n|^2\leqslant (1 +\frac{N + 1}{H})\sum(1-\frac{|h|}{H})\sum a_{n+h}\overline{a}_n
n=1∣h∣<H1⩽n⩽Nn=1|h|<H1\leqslant n\leqslant N
1⩽n+h⩽N 1\leqslant n+h\leqslant N

(“van der Corput inequality”). (Set an=0a_n= 0 if nn is not between 1 and N; write ∑nan=∑nan+h\sum_na_n=\sum_na_{n+h} for 0⩽h<H0\leqslant h <H, sum over hh, then apply the Cauchy-Schwarz inequality.)

Let π:R→R/Z\pi :\mathbf{R}\rightarrow \mathbf{R}/\mathbf{Z} be the canonical projection. We say that a sequence (xn)(x_n) of real numbers is equidistributed modulo 1 if the sequence (νn)(\nu_n) of measures on R/Z\mathbf{R}/\mathbf{Z} defined by

nn

1

νn(f)=∑f(π(xk)),f∈C(R/Z) \nu_n(f) =\sum f(\pi (x_k)),f\in \mathscr{C}(\mathbf{R}/\mathbf{Z})
n n
k=1k=1

converges vaguely towards the Haar measure of total mass 1 on R/Z\mathbf{R}/\mathbf{Z}. b) If, for every integer h⩾1h\geqslant 1, the sequence (xn+h−xn)n(x_{n+h}-x_n)_n is equidistributed modulo 1, then (xn)(x_n) is equidistributed modulo 1. (Use the preceding exercise and the van der Corput inequality.)

c) Let d⩾1d\geqslant 1 be an integer and let P=αdXd+⋯+α0∈R[X]P =\alpha_dX^d+\cdots +\alpha_0\in \mathbf{R}[X] be a polynomial such that αd∉Q\alpha_d\notin\mathbf{Q}. Then the sequence (P(n))(P(n)) is equidistributed modulo 1. (Reason by induction on d⩾1d\geqslant 1.)

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.