Bourbaki

Exercise 30, § 6

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

Spectral Theories, I, p. 185

Bourbaki marks this one as harder than the rest.

Let AA be a Banach algebra having an approximate unit. Denote by A~\widetilde{A} the Banach algebra obtained from AA by adjunction of a unit element, denoted by 1.

a) Let x∈Ax\in A. Then xx belongs to the closed left ideal generated by xx. b) For every ε>0\varepsilon >0 and every finite family (xi)i∈I(x_i)_{i\in I} of elements of A, there exists e∈Ae\in A such that ∥e∥⩽1\|e\|\leqslant 1 and ∥exi−xi∥<ε\|ex_i-x_i\|< \varepsilon for all i∈Ii\in I.

c) Let x∈A,ε>0x\in A,\varepsilon >0 and 0<γ⩽1/80< \gamma \leqslant 1/8. There exists η>0\eta >0 such that for every e∈Ae\in A satisfying ∥e∥⩽1\|e\|\leqslant 1 and ∥ex−x∥<η\|ex-x\|< \eta, then α=(1−γ)⋅1+γe\alpha = (1-\gamma )\cdot 1+\gamma e is invertible in A~\widetilde{A} and ∥x−α−1x∥<ε\|x-\alpha^{-1}x\|< \varepsilon.

d) Let x∈A,ε>0,δ>0x\in A,\varepsilon >0,\delta >0 and 0<γ⩽1/80< \gamma \leqslant 1/8. There exists a sequence (en)n⩾1(e_n)_{n\geqslant 1} in A having the following properties:

(i) We have ∥en∥<2\|e_n\|<2 for all n⩾1n\geqslant 1 ;

(ii) The element

yn=∑k=1nγ(1−γ)k−1ek+(1−γ)n⋅1 y_n=\sum_{k=1}^n\gamma (1-\gamma )^{k-1}e_k+ (1-\gamma )^n\cdot 1

is invertible in A~\widetilde{A} for all n⩾1n\geqslant 1 and

∥yn−1x−yn−1−1x∥<δ2−n(n⩾2),∥y1−1x−x∥<δ/2 \|y_n^{-1}x-y_{n-1}^{-1}x\|< \delta 2^{-n}(n\geqslant 2),\|y^{-1}_1x-x\|< \delta /2

(Assuming that (e1,...,en)(e_1, . . . , e_n) have been defined, consider an arbitrary element en+1e_{n+1} of norm <2<2 ; observe that

yn+1=((1−γ)⋅1+en+1)(∑k=1nγ(1−γ)k−1ek′+(1−γ)n⋅1) y_{n+1}= ((1-\gamma )\cdot 1 +e_{n+1})(\sum_{k=1}^n\gamma (1-\gamma )^{k-1}e'_k+ (1-\gamma )^n\cdot 1)

where ek′=((1−γ)⋅1+en+1)−1eke'_k= ((1-\gamma )\cdot 1 +e_{n+1})^{-1}e_k, and determine en+1e_{n+1} satisfying the required conditions by exploiting the fact that the set of invertible elements of A~\widetilde{A} is open, and that the mapping a↦a−1a\mapsto a^{-1} is continuous.)

e) For every x∈Ax\in A, there exist yy and zz in A such that x=yzx=yz (“Cohen factorization theorem”). Let δ>0\delta >0. We may suppose that zz belongs to the closed left ideal generated by xx and that ∥y−z∥<δ\|y-z\|< \delta. (Consider a sequence (en)n⩾1(e_n)_{n\geqslant 1} as above and define yny_n as previously; the sequence zn=yn−1xz_n=y_n^{-1}x converges to an element z∈Az\in A, and znz_n belongs to the closed left ideal generated by xx; the sequence (yn)(y_n) converges to y∈Ay\in A and x=yz.x=yz.)

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.