Bourbaki

Exercise 57, § 1

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

Spectral Theories, II, p. 294

On retains the notations of the preceding exercise. Let M, N⩾1N\geqslant 1 be integers and let (an)M<n⩽M+N(a_n)_{M<n\leqslant M+N} be a family of complex numbers. Let δ>0\delta >0 be a real number, R⩾1R\geqslant 1 an integer and (ξr)1⩽r⩽R(\xi_r)_{1\leqslant r\leqslant R} a family of real numbers such that infh∈Z∣ξr−ξr′−h∣⩾δ_{h\in\mathbf{Z}}|\xi_r-\xi_{r'}-h|\geqslant \delta for all r≠r′r\not=r'.

For every x∈Rx\in \mathbf{R}, put

S(x)=∑n=M+1M+Nane2iπnx S(x) =\sum_{n=M+1}^{M+N}a_ne^{2i\pi nx}

a) One has

∑r=1R∣S(ξr)∣2⩽(N+δ−1−1)∑n=M+1M+N∣an∣2 \sum_{r=1}^R|S(\xi_r)|^2\leqslant (N +\delta^{-1}-1)\sum_{n=M+1}^{M+N}|a_n|^2

("large sieve inequality").

(Consider the function τ:z↦σ(δz)\tau :z\mapsto \sigma (\delta z), where σ\sigma is the function of questions f) and g) of the preceding exercise for the interval [α,β]=[δ(M+1),δ(M+N)][\alpha , \beta ] = [\delta (M + 1), \delta (M + N)]; put

S∗(x)=∑n=M+1M+Na(n)τ(n)−1e2iπnx S^*(x) =\sum_{n=M+1}^{M+N}a(n)\tau (n)^{-1}e^{2i\pi nx}

and observe that

S(x)=∫−δ/2δ/2τ^(t)S∗(t+x)dt S(x) =\int_{-\delta /2}^{\delta /2}\widehat{\tau}(t)S^*(t+x)dt

Put x=ξrx=\xi_r, apply the Cauchy-Schwarz inequality and sum over rr.)

b) The constant N+δ−1−1N +\delta^{-1}-1 is optimal.

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