Spectral Theories, I, p. 161
Bourbaki marks this one as harder than the rest.
a) Let E be a complex Banach space and x:C−{1}→E an entire function of 1/(ζ−1). We denote by x(ζ)=∑n=0∞anζn,x(ζ)=∑n=0∞bnζ−n the expansions of x for ∣ζ∣<1,∣ζ∣>1 respectively. We suppose that there exists α>0 such that ∥an∥=o(nα),∥bn∥=o(nα). Then x(ζ) is a polynomial in 1/(ζ−1) of degree <α+1. (Let ε>0. We have ∥x(ζ)∥⩽ε(1−r)−1−α for rε⩽r=∣ζ∣<1,∥x(ζ)∥⩽ε(1−r−1)−1−α for 1<r⩽rε′. Putting 1−ζ=(ω+21)−1,ω=Reiθ, now prove that, for 0⩽r<1 and R⩾1, one has
(1−r)−1⩽49 cosRθ
and that, for 1<r and R⩾1,
(1−r−1)−1⩽49 cosRθ.
Put y(ω)=x(ζ) and now prove that R−α−1∣cosθ∣α+1∥y(Reiθ)∥ tends to 0 when R tends to +∞, uniformly in θ. Apply then exerc. 29, b).)
Let A be a complex unital Banach algebra, q∈A a quasi-nilpotent element and x=1+q.
b) In order that qN=0, it is necessary and sufficient that ∥x±n∥=o(nN) when n tends to +∞. (To see that the condition is sufficient, now prove, by applying a), that R(λ,x) is a polynomial in 1/(λ−1) of degree ⩽N. On the other hand, R(λ,x)= ∑n=0∞qn(λ−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∈G, one has ∥xn∥=o(n),∥x−n∥=o(n) for n→+∞, then the restriction to G of the canonical mapping A→A/R is injective.
Nothing.
Nothing.