Bourbaki

Proposition 17

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

Théories spectrales, chapter I, § 6. Algèbres stellaires

Let A be a *-algebra, and let F\mathfrak{F} be a filter basis on A+⩽1A^{\leqslant 1}_+. For F\mathfrak{F} to be an increasing approximate unit of A, it is necessary and sufficient that one have

limf,Ffx=x_{f,\mathfrak{F}}f x=x

for every positive element xx of A.

The condition is obviously necessary; let us prove that it is sufficient. Let A~\widetilde{A} be the *-algebra deduced from A by adjunction of a unit element. Let xx be an element of A and let f∈A+⩽1f\in A^{\leqslant 1}_+. We have

∥fx−x∥2=∥(fx−x)(fx−x)∗∥=∥(f−1)xx∗(f−1)∥ \|f x-x\|^2=\|(f x-x)(f x-x)^*\|=\|(f-1)xx^*(f-1)\|
⩽∥(f−1)xx∗∥ \leqslant \|(f-1)xx^*\|

because ∥f−1∥⩽1\|f-1\|\leqslant 1 (Lemma 12, c) of I, p. 116). We therefore have

lim supf,F∥fx−x∥2⩽_{f,\mathfrak{F}}\|f x-x\|^2\leqslant lim supf,F∥fxx∗−xx∗∥_{f,\mathfrak{F}}\|f xx^*-xx^*\|. Since xx∗xx^* is positive (Theorem 2 of I, p. 118), the assumption implies that

lim supf,F∥fx−x∥2⩽∥xx∗−xx∗∥=0_{f,\mathfrak{F}}\|f x-x\|^2\leqslant \|xx^*-xx^*\|= 0,

so that

limf,Ffx=x_{f,\mathfrak{F}}f x=x.

Since the involution of A is continuous and since F\mathfrak{F} is a filter basis on AhA_h, we have

limf,Fxf=_{f,\mathfrak{F}}xf= limf,F(f∗x∗)∗=_{f,\mathfrak{F}}(f^*x^*)^*= limf,F(fx∗)∗=(x∗)∗=x_{f,\mathfrak{F}}(f x^*)^*= (x^*)^*=x.

The proposition follows.

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-5-6-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.