Bourbaki

Exercise 1, § 1

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

Spectral Theories, II, p. 262

a) Let ω,p,q∈R\omega , p, q\in \mathbf{R} with p⩽qp\leqslant q. Put f(x)=eiωxf(x) =e^{i\omega x} for p⩽x⩽qp\leqslant x\leqslant q, f(x)=0f(x) = 0 for x>qx > q or x<px < p. Then for y∈Ry\in \mathbf{R}, one has

F(f)(y)=ieip(ω−2πy)−eiq(ω−2πy)ω−2πy. \mathscr{F}(f)(y) =i\frac{e^{ip(\omega-2\pi y)}-e^{iq(\omega-2\pi y)}}{\omega -2\pi y}.

b) Let ω∈R,β>0\omega \in \mathbf{R},\beta >0. Put f(x)=e(−β+iω)xf(x) =e^{(-\beta+i\omega)x} for x⩾0,f(x)=0x\geqslant 0,f(x) = 0 for x<0x <0. Then for y∈Ry\in \mathbf{R}, one has

F(f)(y)=iω−2πy+iβ. \mathscr{F}(f)(y) =\frac{i}{\omega -2\pi y+i\beta}.

c) Let a>0a >0. Put f(x)=1ae−a∣x∣f(x) =\frac{1}{a}e^{-a|x|} for x∈Rx\in \mathbf{R}. Then for y∈Ry\in \mathbf{R}, one has

F(f)(y)=24π2y2+a2. \mathscr{F}(f)(y) =\frac{2}{4\pi^2y^2+a^2}.

d) Put f(x)=e−πx2f(x) =e^{-\pi x^2} for x∈Rx\in \mathbf{R}. Then for y∈Ry\in \mathbf{R}, one has F(f)(y)=\mathscr{F}(f)(y) = e−πy2e^{-\pi y^2}. (Find a first-order linear differential equation satisfied

by the Fourier transform of ff, and use the formula ∫0+∞e−t2dt=12π\int_0^{+\infty}e^{-t^2}dt=\frac{1}{2}\sqrt{\pi} of FVR, VII, §1, No.o3.)^o3.)

e) Let n∈Nn\in \mathbf{N} and let Q be a positive definite quadratic form on Rn\mathbf{R}^n. Let σ\sigma be a linear mapping on Rn\mathbf{R}^n such that Q(x)=∥σ(x)∥2Q(x) =\|\sigma (x)\|^2. Let Q∗Q^* be the positive definite quadratic form such that Q∗(y)=∥tσ−1(y)∥2Q^*(y) =\|^t\sigma^{-1}(y)\|^2 for every y∈Rny\in \mathbf{R}^n. Let z∈Cz\in \mathbf{C} be a complex number such that I(z)>0\mathscr{I}(z)>0. The mapping ff of Rn\mathbf{R}^n into C\mathbf{C} defined by f(x)=f(x) = exp(iπzQ(x)i\pi zQ(x)) belongs to S(Rn)\mathscr{S}(\mathbf{R}^n) and one has

F(f)(y)=∣det⁡(σ)∣−1(iz)n/2exp⁡(iπz−1Q∗(y)) \mathscr{F}(f)(y) =|\det(\sigma )|^{-1}\left(\frac{i}{z}\right)^{n/2}\exp(i\pi z^{-1}Q^*(y))

for every y∈Rny\in \mathbf{R}^n.

f) Let f∈C(R)f\in \mathscr{C}(\mathbf{R}) be the function defined by f(x)=f(x) = (sin(πx\pi x)/(πx)/(\pi x))2^2 for x≠0x\not= 0 and f(0)=1f(0) = 1. We have F(f)(y)=0\mathscr{F}(f)(y) = 0 if ∣y∣>1|y|>1 and F(f)(y)=1−∣y∣\mathscr{F}(f)(y) = 1- |y| otherwise. Let g(x)=xf(x)g(x) =xf(x). We have F(g)(y)=0\mathscr{F}(g)(y) = 0 if ∣y∣>1|y|>1 and F(g)(y)=\mathscr{F}(g)(y) = s(yy)/(2iπ)/(2i\pi ) otherwise.

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Machine translation of the English, by gpt-5.4. Not checked by a person.