Bourbaki

Lemma 7

Tag 02DC, permanent. Cite it as [Bourbaki, Tag 02DC].

Théories spectrales, chapter I, § 6. Algèbres stellaires

Let A be a stellar algebra.

a) The regular representation γ\boldsymbol{\gamma } of A (def. 1 of I, p. 16) is isometric, that is to say

∥x∥=\|x\|= sup∥y∥⩽1∥xy∥_{\|y\|\leqslant 1}\|xy\|,

for all x∈Ax\in A;

b) For all x∈Ahx\in A_h, one has

ϱ(x)=∥x∥(2) \varrho (x) =\|x\| \tag{2}

Let x∈Ax\in A. One has sup∥y∥⩽1∥xy∥⩽∥x∥_{\|y\|\leqslant 1}\|xy\|\leqslant \|x\|. To prove that ∥x∥⩽\|x\|\leqslant sup∥y∥⩽1∥xy∥_{\|y\|\leqslant 1}\|xy\|, one may suppose that ∥x∥=1\|x\|= 1. Then the element yy = x∗x^* satisfies ∥y∥=∥x∗∥=1\|y\|=\|x^*\|= 1, and ∥xy∥=∥x∥2=1\|xy\|=\|x\|^2= 1, whence assertion a).

Suppose that xx is hermitian. Then ∥x2∥=∥x∗x∥=∥x∥2\|x^2\|=\|x^*x\|=\|x\|^2, whence by induction ∥x2n∥2−n=∥x∥\|x^{2^n}\|^{2^{-n}}=\|x\| for every integer n⩾1n\geqslant 1, whence assertion b) by prop. 1 of I, p. 20.

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-5-6-mini, gpt-5.4, 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.