Bourbaki

Lemma 2

Tag 0CAE, permanent. Cite it as [Bourbaki, Tag 0CAE].

General Topology, chapter IX, Appendix 0. Infinite products in normed algebras

If II is a finite subset of N\mathbf{N} such that φ(I)<1/c\varphi(I) < 1/c, then

σI≤cφ(I)/(1−cφ(I)). \sigma_I \leq c\varphi(I)/(1 - c\varphi(I)).

For since σJ≤σI\sigma_J \leq \sigma_I for every subset JJ of II, we have from (2)

∣∣sJ∣∣≤φ(I)σI+∣∣pJ−e∣∣≤(1+σI)φ(I); ||s_J|| \leq \varphi(I)\sigma_I + ||p_J - e|| \leq (1 + \sigma_I)\varphi(I);

and since also σI≤c⋅sup⁡J⊂I∣∣sJ∣∣\sigma_I \leq c \cdot \sup_{J \subset I} ||s_J||, from (1), it follows that

σI≤cφ(I)(1+σI), \sigma_I \leq c\varphi(I)(1 + \sigma_I),

which leads to the result.

Now let (e+un)(e + u_n) be a multipliable sequence in AA, whose product is a unit; by Theorem 1 there exists a finite subset J0J_0 of N\mathbf{N} such that, for each finite subset HH of N\mathbf{N} which does not meet J0J_0, we have

∣∏i∈H(e+ui)−e∣≤1/2c. \left| \prod_{i \in H} (e + u_i) - e \right| \leq 1/2c.

By Lemma 2, it follows that ∑i∈H∣∣ui∣∣−1\sum_{i \in H} ||u_i|| - 1 for every finite subset HH of N\mathbf{N} which does not meet J0J_0, and hence (Chapter IV, § 7, no. 1, Theorem 1) the family (∣∣un∣∣)(||u_n||) is summable in R\mathbf{R}.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.