Bourbaki

Remark

Tag 02K9, permanent. Cite it as [Bourbaki, Tag 02K9].

Théories spectrales, chapter II, § 1. Transformation de Fourier

The elementary formulas concerning the Fourier transform of measures remain valid for the Fourier transform of tempered distributions.

Thus, if α∈Nn\alpha \in \mathbf{N}^n and f∈S′(Rn)f\in \mathscr{S}'(\mathbf{R}^n), one has

F(∂αf)=(2iπ)∣α∣XαF(f) \mathscr{F}(\partial^{\alpha}f) = (2i\pi )^{|\alpha|}X^{\alpha}\mathscr{F}(f)
F(Xαf)=(−2iπ)−∣α∣∂α(F(f)) \mathscr{F}(X^{\alpha}f) = (-2i\pi )^{-|\alpha|}\partial^{\alpha}(\mathscr{F}(f))

Let y∈Rny\in \mathbf{R}^n. Let us denote by γ(y)\boldsymbol{\gamma }(y) the endomorphism of S′(Rn)\mathscr{S}'(\mathbf{R}^n) defined by

⟨γ(y)f,φ⟩=⟨f,γ(−y)φ⟩ \langle \boldsymbol{\gamma }(y)f, \varphi \rangle =\langle f,\boldsymbol{\gamma }(-y)\varphi \rangle

for f∈S′(Rn)f\in \mathscr{S}'(\mathbf{R}^n) and φ∈S(Rn)\varphi \in \mathscr{S}(\mathbf{R}^n). Let eye_y be the character of Rn\mathbf{R}^n such that ey(x)=e_y(x) = exp(2iπx⋅yi\pi x\cdot y). Then ey∈S′(Rn)e_y\in \mathscr{S}'(\mathbf{R}^n). We have F(ey)=εy\mathscr{F}(e_y) =\varepsilon_y, and more generally

F(eyf)=γ(y)F(f) \mathscr{F}(e_yf) =\boldsymbol{\gamma }(y)\mathscr{F}(f)

for every f∈S′(Rn).∗f\in \mathscr{S}'(\mathbf{R}^n).*

Let n⩾1n\geqslant 1 be an integer, and let G=TnG =\mathbf{T}^n, endowed with the normalized Haar measure. The dual group of G is identified with Zn\mathbf{Z}^n by the mapping h↦χhh\mapsto \chi_h, where χh\chi_h is the unitary character of Tn\mathbf{T}^n obtained by passing to the quotient from the character x↦x\mapsto exp(2iπh⋅xi\pi h\cdot x) of Rn\mathbf{R}^n (Corollary 3 of II, p. 236). The Fourier transform of a measure μ\mu on Tn\mathbf{T}^n is identified with the family (μ^(h))h∈Zn(\widehat{\mu}(h))_{h\in\mathbf{Z}^n} where

μ^(h)=∫Tne−2iπh⋅xdμ(x) \widehat{\mu}(h) =\int_{\mathbf{T}^n}e^{-2i\pi h\cdot x}d\mu(x)

The series

∑h∈Znμ^(h)χh \sum_{h\in\mathbf{Z}^n}\widehat{\mu}(h)\chi_h

is called the Fourier series of μ\mu.

If f∈L1(Tn)f\in L^1(\mathbf{T}^n) is such that its Fourier series converges absolutely in L1(Zn)L^1(\mathbf{Z}^n), then f∈C(Tn)f\in \mathscr{C}(\mathbf{T}^n) and

f(x)=∑h∈Znf^(h)e2iπh⋅x f(x) =\sum_{h\in\mathbf{Z}^n}\widehat{f}(h)e^{2i\pi h\cdot x}

for every x∈Tnx\in \mathbf{T}^n (Theorem 3 of II, p. 222), where

f^(h)=∫Tnf(x)e−2iπh⋅xdx,h∈Zn \widehat{f}(h) =\int_{\mathbf{T}^n}f(x)e^{-2i\pi h\cdot x}dx,h\in \mathbf{Z}^n

For f∈L2(Tn)f\in L^2(\mathbf{T}^n), the fourier inversion formula (Proposition 12 of II, p. 219) says that, if the series with general term f^(h)\widehat{f}(h) converges absolutely, one has

f(x)=∑h∈Znf^(h)e2iπh⋅x f(x) =\sum_{h\in\mathbf{Z}^n}\widehat{f}(h)e^{2i\pi h\cdot x}

for almost every xx in Tn\mathbf{T}^n.

However, even if ff is continuous, the Fourier series of ff does not generally converge to f(x)f(x) for every xx (Exercise 30 of II, p. 274). The following result is all the more useful.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Machine translation of the English, by gpt-5.4, gpt-5-6-mini, against glossary version 34. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.