Bourbaki

Exercise 66, § 1

Tag 02M6, permanent. Cite it as [Bourbaki, Tag 02M6].

Spectral Theories, II, p. 300

Bourbaki marks this one as harder than the rest.

We denote by e:T→C∗e:\mathbf{T}\rightarrow \mathbf{C}^* the character defined by passing to the quotient by the character x↦e2iπxx\mapsto e^{2i\pi x} of R\mathbf{R}.

Let d⩾2d\geqslant 2 be an integer. Let α∈R\alpha \in \mathbf{R} and let g∈R[X]g\in \mathbf{R}[X] be a polynomial whose degree is <d< d. Let aa and qq be relatively prime integers, with q≠0q\not= 0, such that

∣α−a∣⩽12 |\alpha -a|\leqslant 1_2
qq qq

a) Suppose that α∉Z\alpha \notin \mathbf{Z}. Let N and M be integers with M⩾0M\geqslant 0. We have

∣∑N+Me2iπαn∣⩽1∣∣ |\sum^{N+M}e^{2i\pi \alpha n}|\leqslant \frac{1}{||}

sin(πα\pi \alpha )

n=Nn=N

b) We denote D=2d−1D = 2^{d-1}. For every ε>0\varepsilon >0, there exists a real number C(ε,d)⩾0C(\varepsilon , d)\geqslant 0 such that for every integer P⩾1P\geqslant 1, one has

∣∑Pe2iπ(αnd+g(n))∣⩽C(ε,d)P1+ε(P−1/D+q−1/D+(Pdq)−1/D) |\sum^Pe^{2i\pi(\alpha n^d+g(n))}|\leqslant C(\varepsilon , d)P^{1+\varepsilon}(P^{-1/D}+q^{-1/D}+(\frac{P^d}{q})^{-1/D})
n=1n=1

(Proceed by induction on dd by reducing the square of the left-hand term of the inequality to a family of polynomials of degree ⩽d−1\leqslant d-1; when d=2d= 2, apply then question a). One may prove and use the estimate

d(n)=O(nε) d(n) = O(n^{\varepsilon})

when n→+∞n\rightarrow +\infty, where n⩾1n\geqslant 1 and d(n)d(n) is the number of divisors of nn in N\mathbf{N}, valid for every real number ε>0\varepsilon >0.)

c) Deduce a new proof of the assertion of question c) of exercise 35.

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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.