Bourbaki

Lemma 6

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

Théories spectrales, chapter V, § 2. Représentations des groupes localement compacts

Let f∈Kπ‾(G)f\in \mathscr{K}_{\overline{\pi}}(G). For every g∈Gg\in G, the function

f~:x↦η(g−1,xH)1/pf(g−1x) \widetilde{f}:x\mapsto \eta (g^{-1}, xH)^{1/p}f(g^{-1}x)

of G into E belongs to Kπ‾(G)\mathscr{K}_{\overline{\pi}}(G) and satisfies Np(f~)=Np(f)N_p(\widetilde{f}) = N_p(f).

It is verified without difficulty that f~∈Kπ‾(G)\widetilde{f}\in \mathscr{K}_{\overline{\pi}}(G). Since

Np(f~)p=∫G/H∗∥f~∥pdν=∫G/H∗γG/H(g)(∥f∥p)(y)η(g−1,y)dν(y) N_p(\widetilde{f})^p=\int_{G/H}^*\|\widetilde{f}\|^pd\nu =\int_{G/H}^*\gamma_{G/H}(g)(\|f\|^p)(y)\eta (g^{-1}, y)d\nu (y)

and since (y↦η(g−1,y))⋅ν=γG/H(g)ν(y\mapsto \eta (g^{-1}, y))\cdot \nu =\boldsymbol{\gamma }_{G/H}(g)\nu, one obtains Np(f~)=Np(f)N_p(\widetilde{f}) = N_p(f).

It follows from this lemma that there exists a continuous isometric representation π~\widetilde{\pi} of G in Lπ‾p(G)L^p_{\overline{\pi}}(G) such that for f∈Kπ(G)f\in \mathscr{K}_{\pi}(G) and g∈Gg\in G, the element π~(g)f\widetilde{\pi}(g)f is the class of the function f~\widetilde{f} defined above. If p=2p= 2, then this representation is unitary.

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.

Cited by

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Cites

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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.