Bourbaki

Exercise 27, § 2

Tag 030A, permanent. Cite it as [Bourbaki, Tag 030A].

Spectral Theories, IV, p. 328

Bourbaki marks this one as harder than the rest.

The object is to prove the converse of the assertion in question b) of exercise 25. For this one will use the results and the notations of exercise 23 of V, p. 503 below. In particular, G denotes the group R+∗\mathbf{R}^*_+ endowed with the Haar measure μ=x−1dx\mu=x^{-1}dx, dual to the group iRi\mathbf{R} by the mapping (x,it)↦xit(x, it)\mapsto x^{it}. Denote by X the identity mapping of G.

Denote by co (resp. si, resp. ta) the function it↦it\mapsto cos(iπti\pi t) (resp. the function it↦it\mapsto sin(iπti\pi t), the function it↦it\mapsto tan(iπti\pi t)) on iRi\mathbf{R}.

Let f∈f\in Loe(G) satisfy conditions (i), (ii) and (iii) of question f ) of exercise 25.

a) Let α\alpha be the function x↦x/(1+x)x\mapsto x/(1 +x) on G. One has Iα=]−1,0[I_{\alpha}= ]-1,0[ and α^(s)=\widehat{\alpha}(s) = −π/-\pi / sin(πs\pi s) for s∈Bfs\in B_f.

b) The function ta is bounded on iRi\mathbf{R}. Denote by v=F∘mta∘F‾v=\mathscr{F}\circ m_{ta}\circ \overline{\mathscr{F}}; this is an endomorphism of L2(G)L^2(G) such that ∥v∥⩽1\|v\|\leqslant 1 and R(v)=0\mathscr{R}(v) = 0.

c) Let ε>0\varepsilon >0 be a real number. Denote by uεu_{\varepsilon} the endomorphism of multiplication by hε=X/(1+εX)h_{\varepsilon}= X/(1 +\varepsilon X) on L2(G)L^2(G). One has uε⩾0u_{\varepsilon}\geqslant 0 and for every φ∈L2(G)\varphi \in L^2(G) and every bounded continuous function ff on G, one has

1

⟨φ∣R(f(uε)v)φ⟩=⟨f∘hε∣Q(φ)⟩ \langle \varphi |\mathscr{R}(f(u_{\varepsilon})v)\varphi \rangle =\langle f\circ h_{\varepsilon}|Q(\varphi )\rangle

2

where Q(φ)=φv(φ)+v(φ)‾φQ(\varphi ) =\varphi v(\varphi ) +\overline{v(\varphi)}\varphi.

d) Let F be the vector subspace of L2(iR)L^2(i\mathbf{R}) generated by the functions t↦eα(it−s0)2t\mapsto e^{\alpha(it-s_0)^2} where α∈R+∗\alpha \in \mathbf{R}_+^* and s0∈Cs_0\in \mathbf{C}; it is dense in L2(iR)L^2(i\mathbf{R}). e) Let ψ∈F\psi \in F. There exists φ∈L2(G)\varphi \in L^2(G) such that φ^=\widehat{\varphi}= co ⋅ψ\cdot \psi. Let g=Q(φ)g= Q(\varphi ). One has g∈L1g\in L^1(G), and there exists a holomorphic function on C\mathbf{C} whose restriction to iRi\mathbf{R} coincides with g^\widehat{g}. Moreover

g^=12iπ(ψ∗∗ψ)\widehat{g}=\frac{1}{2i\pi}(\psi^**\psi ) si.

f) For every ε>0\varepsilon >0, one has R(uε∘v)⩾0\mathscr{R}(u_{\varepsilon}\circ v)\geqslant 0. (Calculate ⟨φ∣R(uεu)φ⟩\langle \varphi |\mathscr{R}(u_{\varepsilon}u)\varphi \rangle when φ\varphi is a function as in the preceding question, and use the convolution on G as in question d) of exercise 23 of V, p. 503.)

g) Let f∈f\in Loe(R+∗\mathbf{R}^*_+). Let gg be a function associated with ψ∈F\psi \in F as in question e). One then has

∫Gfgdμ⩾0 \int_Gf g d\mu\geqslant 0

and

∫Rf^(−σ−it)\int_{\mathbf{R}}\widehat{f}(-\sigma -it) sin(π(σ+it)\pi (\sigma +it))(ψ∗∗ψ)(σ+it)dt⩾0(\psi^**\psi )(\sigma +it)dt\geqslant 0

for every σ∈]0,1[\sigma \in ]0,1[.

h) Let σ∈]0,1[\sigma \in ]0,1[. The function qq defined by q(s)=f^(−s)q(s) =\widehat{f}(-s) sin(πs\pi s) for ss such that R(s)=σ\mathscr{R}(s) =\sigma satisfies

nnnn ∑∑t‾itjq(12(z‾i+zj))⩾0\sum\sum\overline{t}_it_jq(\frac{1}{2}(\overline{z}_i+z_j))\geqslant 0 i=0j=0i=0j=0

for every n∈Nn\in \mathbf{N} and every families (ti)0⩽i⩽n(t_i)_{0\leqslant i\leqslant n} and (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 every ii.

i) There exists a positive measure ν0\nu_0 on G such that ν^0(it)=f^(−it)\widehat{\nu}_0(it) =\widehat{f}(-it) sin(iπti\pi t)/π/\pi for every t∈Rt\in \mathbf{R}; one has Xσ∈L1(G,ν0)X^{\sigma}\in L^1(G, \nu_0) if 0<σ<10< \sigma <1.

j) Let ν=νˇ0\nu =\check{\nu}^0. The measure ν\nu is bounded. (Verify that]−1,0[⊂Iν]-1,0[\subset I_{\nu} and then that ν^(−t)\widehat{\nu}(-t) is bounded when t→0t\rightarrow 0 in order to prove that ν([1,+∞[)\nu ([1,+\infty [) is finite.) For every xx, one has

f(x)=∫Gxx+tdν(t) f(x) =\int_G\frac{x}{x + t}d\nu (t)
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