Bourbaki

Exercise 23, § 3

Tag 03DM, permanent. Cite it as [Bourbaki, Tag 03DM].

Spectral Theories, V, p. 503

Let G=R+∗G =\mathbf{R}^*_+; it is a locally compact commutative group and the measure μ=dx/x\mu=dx/x is a Haar measure on G. The group G and the group iRi\mathbf{R} are in duality by the mapping (x,it)↦xit(x, it)\mapsto x^{it}, the dual measure of μ\mu being the Lebesgue measure on iR(i\mathbf{R}(cf. cor. 4 of II, p. 236). We denote by F\mathscr{F} the Fourier transform of G and by F‾\overline{\mathscr{F}} the Fourier cotransform of G^\widehat{G}.

We denote by X the identity function of G.

a) Let ν\nu be a positive measure on G. We denote by IνI_{\nu} the set of σ∈R\sigma \in \mathbf{R} such that the function XσX^{\sigma} is ν\nu-integrable, and by BνB_{\nu} the set of s∈Rs\in \mathbf{R} such that R(s)∈Iν\mathscr{R}(s)\in I_{\nu}. The set IνI_{\nu} is an interval of R\mathbf{R} and the function s↦ν(Xs)s\mapsto \nu (X^s) is defined and holomorphic in the interior of BνB_{\nu}.

b) Let ff be a locally integrable function on G. We denote by If=If⋅μI_f= I_{f\cdot\mu} and Bf=Bf⋅μB_f= B_{f\cdot\mu}. For every s∈Bfs\in B_f, we put

f^(s)=∫Gf(x)xsdμ(x) \widehat{f}(s) =\int_Gf(x)x^sd\mu(x)

If there exist δ>0\delta >0 and c⩾0c\geqslant 0 such that ∣f^(s)∣⩽c(1+∣s∣)−1−δ|\widehat{f}(s)|\leqslant c(1 +|s|)^{-1-\delta} for s∈Bfs\in B_f, then for every σ\sigma in the interior of BfB_f, one has

f(x)=12π∫Rf^(σ+it)x−σ−itdt f(x) =\frac{1}{2\pi}\int_{\mathbf{R}}\widehat{f}(\sigma +it)x^{-\sigma-it}dt

c) Let f∈L2(G)f\in \mathscr{L}^2(G). Suppose that 0 belongs to the interior of IfI_f. One then has F‾f(it)=f^(it)\overline{\mathscr{F}}f(it) =\widehat{f}(it) for almost all t∈Gt\in G. d) Let ff and gg be measurable functions on G. Let σ\sigma be a real number such that X−2σf∈L2(G)X^{-2\sigma}f\in L^2(G) and X2σg∈L2(G)X^{2\sigma}g\in L^2(G). One then has

∫Gfgdμ=12π∫Rf^(−σ−it)g^(σ+it)dt \int_Gf g d\mu=\frac{1}{2\pi}\int_{\mathbf{R}}\widehat{f}(-\sigma -it)\widehat{g}(\sigma +it)dt

e) Let ff and gg be measurable functions on G. Suppose that

(i) There exists c∈Rc\in \mathbf{R} such that ∣f(x)∣⩽cx/(1+x)|f(x)|\leqslant c x/(1 +x) for every x∈Gx\in G;

(ii) One has]0,1[⊂Ig]0,1[\subset I_g and for every σ\sigma such that 0<σ<10< \sigma <1, the function XσgX^{\sigma}g is bounded ;

(iii) There exists c′∈Rc'\in \mathbf{R} such that ∣g^(s)∣⩽c′(1+∣s∣)−2|\widehat{g}(s)|\leqslant c'(1 +|s|)^{-2} for every s∈Cs\in \mathbf{C} such that R(s)∈]0,1[\mathscr{R}(s)\in ]0,1[.

Then one has

∫Gfgdμ=12π∫Rf^(−σ−it)g^(σ+it)dt \int_Gf g d\mu=\frac{1}{2\pi}\int_{\mathbf{R}}\widehat{f}(-\sigma -it)\widehat{g}(\sigma +it)dt

for every σ∈R\sigma \in \mathbf{R} such that 0<σ<10< \sigma <1.

f) Let σ∈R\sigma \in \mathbf{R} and let f∈C(σ+iR)f\in \mathscr{C}(\sigma +i\mathbf{R}). There exists a positive measure ν\nu on G such that σ∈Iν\sigma \in I_{\nu} and ν^(σ+it)=f(σ+it)\widehat{\nu}(\sigma +it) =f(\sigma +it) for all t∈Rt\in \mathbf{R} if and only if for every integer n⩾1n\geqslant 1, every family (ti)0⩽i⩽n(t_i)_{0\leqslant i\leqslant n} of complex numbers and every family (zi)0⩽i⩽n(z_i)_{0\leqslant i\leqslant n} of complex numbers such that R(zi)=σ\mathscr{R}(z_i) =\sigma for all ii, one has

nnnn
(1) (1)
∑∑t‾itjf(zi+zj)⩾0 \sum\sum\overline{t}_it_jf(z_i+z_j)\geqslant 0

2

i=0j=0i=0j=0
Solution

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