Bourbaki

Lemma 4

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

Théories spectrales, chapter V, § 3. Fonctions de type positif

Let φ∈\varphi \in Pos(G). The function φ\varphi is bounded on G and one has ∥φ∥∞=φ(e)\|\varphi \|_{\infty}=\varphi (e). Moreover, one has

∣φ(g−1h)−φ(h)∣⩽√2φ(e)(φ(e)−R(φ(g)))‾(1) |\varphi (g^{-1}h)-\varphi (h)|\leqslant \surd\overline{2\varphi(e)(\varphi(e) -\mathscr{R}(\varphi(g)))} \tag{1}

for every (g,h)∈G×G(g, h)\in G\times G.

Let (ϱ,x)(\varrho , x) be a Hilbert realization of φ\varphi. One has φ(e)=∥x∥2\varphi (e) =\|x\|^2. For every g∈Gg\in G, one therefore has ∣φ(g)∣=∣⟨x∣ϱ(g)x⟩∣⩽φ(e)|\varphi (g)|=|\langle x|\varrho (g)x\rangle |\leqslant \varphi (e) by the Cauchy-Schwarz inequality. This proves the first assertion.

For every (g,h)∈G×G(g, h)\in G\times G, one has

φ(g−1h)−φ(h)=⟨x∣ϱ(g−1h)x⟩−⟨x∣ϱ(h)x⟩=⟨ϱ(g)x−x∣ϱ(h)x⟩ \varphi (g^{-1}h)-\varphi (h) =\langle x|\varrho (g^{-1}h)x\rangle - \langle x|\varrho (h)x\rangle =\langle \varrho (g)x-x|\varrho (h)x\rangle

since ϱ\varrho is unitary, whence

∣φ(g−1h)−φ(h)∣⩽∥ϱ(g)x−x∥∥ϱ(h)x∥⩽φ(e)1/2∥ϱ(g)x−x∥ |\varphi (g^{-1}h)-\varphi (h)|\leqslant \|\varrho (g)x-x\|\|\varrho (h)x\|\leqslant \varphi (e)^{1/2}\|\varrho (g)x-x\|

We conclude by observing that

∥ϱ(g)x−x∥2=2∥x∥2−2R(⟨x∣ϱ(g)x⟩)=2(φ(e)−R(φ(g))) \|\varrho (g)x-x\|^2= 2\|x\|^2-2\mathscr{R}(\langle x|\varrho (g)x\rangle ) = 2(\varphi (e)-\mathscr{R}(\varphi (g)))

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.