Bourbaki

Exercise 30, § 2

Tag 027H, permanent. Cite it as [Bourbaki, Tag 027H].

Spectral Theories, I, p. 161

Bourbaki marks this one as harder than the rest.

a) Let EE be a complex Banach space and x:C−{1}→Ex:\mathbf{C}-\{1\} \rightarrow E an entire function of 1/(ζ−1)1/(\zeta -1). We denote by x(ζ)=∑n=0∞anζn,x(ζ)=∑n=0∞bnζ−nx(\zeta ) =\sum^{\infty}_{n=0}a_n\zeta^n,x(\zeta ) =\sum^{\infty}_{n=0}b_n\zeta^{-n} the expansions of xx for ∣ζ∣<1,∣ζ∣>1|\zeta |<1,|\zeta |>1 respectively. We suppose that there exists α>0\alpha >0 such that ∥an∥=o(nα),∥bn∥=o(nα)\|a_n\|=o(n^{\alpha}),\|b_n\|=o(n^{\alpha}). Then x(ζ)x(\zeta ) is a polynomial in 1/(ζ−1)1/(\zeta -1) of degree <α+1< \alpha + 1. (Let ε>0\varepsilon >0. We have ∥x(ζ)∥⩽ε(1−r)−1−α\|x(\zeta )\|\leqslant \varepsilon (1-r)^{-1-\alpha} for rε⩽r=∣ζ∣<1,∥x(ζ)∥⩽ε(1−r−1)−1−αr_{\varepsilon}\leqslant r=|\zeta |<1,\|x(\zeta )\|\leqslant \varepsilon (1-r^{-1})^{-1-\alpha} for 1<r⩽rε′1< r\leqslant r'_{\varepsilon}. Putting 1−ζ=(ω+12)−1,ω=Reiθ1-\zeta = (\omega +\frac{1}{2})^{-1},\omega = Re^{i\theta}, now prove that, for 0⩽r<10\leqslant r <1 and R⩾1R\geqslant 1, one has

(1−r)−1⩽94(1-r)^{-1}\leqslant \frac{9}{4} cosRθR\theta

and that, for 1<r1< r and R⩾1R\geqslant 1,

(1−r−1)−1⩽94(1-r^{-1})^{-1}\leqslant \frac{9}{4} cosRθR\theta.

Put y(ω)=x(ζ)y(\omega ) =x(\zeta ) and now prove that R−α−1∣R^{-\alpha-1}|cosθ∣α+1∥y(Reiθ)∥\theta |^{\alpha+1}\|y(Re^{i\theta})\| tends to 0 when R tends to +∞+\infty, uniformly in θ\theta. Apply then exerc. 29, b).)

Let A be a complex unital Banach algebra, q∈Aq\in A a quasi-nilpotent element and x=1+qx= 1 +q.

b) In order that qN=0q^N= 0, it is necessary and sufficient that ∥x±n∥=o(nN)\|x^{\pm n}\|=o(n^N) when nn tends to +∞+\infty. (To see that the condition is sufficient, now prove, by applying a), that R(λ,x)R(\lambda , x) is a polynomial in 1/(λ−1)1/(\lambda -1) of degree ⩽N\leqslant N. On the other hand, R(λ,x)=R(\lambda , x) = ∑n=0∞qn(λ−1)−n−1\sum^{\infty}_{n=0}q^n(\lambda -1)^{-n-1}.)

c) Let R be the radical of A and G a subgroup of the group of invertible elements of A. If, for every x∈Gx\in G, one has ∥xn∥=o(n),∥x−n∥=o(n)\|x^n\|=o(n),\|x^{-n}\|=o(n) for n→+∞n\rightarrow +\infty, then the restriction to G of the canonical mapping A→A/RA\rightarrow A/R is injective.

Solution

No solution has been written yet.

Cited by

Nothing.

Cites

Nothing.

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.