Bourbaki

Lemma 7

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

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

Let ψ∈L1(G)\psi \in L^1(G) be such that ψ⩾0\psi \geqslant 0 and ∫ψ=1\int\psi = 1. Let us denote by pp the seminorm on Cb(G)\mathscr{C}_b(G) defined by p(φ)=∣⟨φ,ψ⟩∣p(\varphi ) =|\langle \varphi , \psi \rangle | for all φ∈Cb(G)\varphi \in \mathscr{C}_b(G). For every φ∈\varphi \in Pos1(G)_1(G), one has

∥ψ∗φ−φ∥∞⩽√2p(1−φ)‾ \|\psi *\varphi -\varphi \|_{\infty}\leqslant \surd\overline{2p(1 -\varphi)}

Let φ∈\varphi \in Pos1(G)_1(G) and x∈Gx\in G. By INT, VIII, p. 167, § 4, no5^o5, prop. 14, we obtain

∣ψ∗φ(x)−φ(x)∣=∣∫G(φ(y−1x)−φ(x))ψ(y)dμ(y)∣ |\psi *\varphi (x)-\varphi (x)|=|\int_G(\varphi (y^{-1}x)-\varphi (x))\psi (y)d\mu(y)|
⩽∫G∣φ(y−1x)−φ(x)∣ψ(y)dμ(y) \leqslant \int_G|\varphi (y^{-1}x)-\varphi (x)|\psi (y)d\mu(y)
⩽∫G√2(1−Rφ(y))‾ψ(y)dμ(y) \leqslant \int_G\surd\overline{2(1 -\mathscr{R}\varphi(y))}\psi (y)d\mu(y)

applying the majorization (1) of V, p. 446. The Cauchy-Schwarz inequality then implies

√∫1/2∫1/2\surd \int 1/2\int 1/2
∥ψ∗φ−φ∥∞⩽2‾(G(1−R(φ))ψdμ)(Gψdμ) \|\psi *\varphi -\varphi \|_{\infty}\leqslant \overline{2}(_G(1-\mathscr{R}(\varphi ))\psi d\mu)(_G\psi d\mu)
√√ \surd \surd

⩽\leqslant 2 p(1−φ)p(1-\varphi ),

whence the result.

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

Nothing.

Cites

Nothing.

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.