Let μ be an isolated point of SpA(x). Then
a) For every λ∈C− SpA(x), we have
R(x,λ)=R{μ}(x,λ)+RSpA(x)−{μ}(x,λ) ;
b) The function which associates to λ RSpA(x)−{μ}(x,λ) is holomorphic in C− SpA(x) and in a neighbourhood of μ; moreover, the function which associates to λ R{μ}(x,λ) is holomorphic in C−{μ};
c) One has
limn→+∞∥(x−μ)nj{μ}∥1/n=0
and, for λ∈C−{μ}, the formula
R{μ}(x,λ)=n=0∑∞(λ−μ)−n−1(x−μ)nj{μ}(20)
The foregoing implies assertions a) and b). Let us prove c). Replacing x by x−μ, we are reduced to the case where μ=0. Put H={0}; this is an open and closed subset of SpA(x). By formula (18), the spectrum of xH in AH is {0}, hence xH is quasi-nilpotent, that is to say, ∥xnjH∥1/n=∥(xjH)n∥1/n tends to 0 as n tends to +∞. Moreover, for λ=0, one has in AH
(λjH−xH)−1=n=0∑∞λ−n−1xHn
(Theorem 1 of I, p. 24, d)), whence (20).
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.
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