Bourbaki

Exercise 28, § 2

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

Spectral Theories, I, p. 160

Let AA be a unital complex Banach algebra, Δ\Delta an open subset of C\mathbf{C} and λ↦R(λ)\lambda \mapsto R(\lambda ) a mapping from Δ\Delta into AA such that

R(λ)−R(μ)=−(λ−μ)R(λ)R(μ)(5) R(\lambda )-R(\mu) =-(\lambda -\mu)R(\lambda )R(\mu) \tag{5}

for all λ,μ∈Δ\lambda , \mu\in \Delta. (a) Let λ0∈Δ\lambda_0\in \Delta. For all λ∈Δ\lambda \in \Delta such that ∣λ−λ0∣⋅∥R(λ)∥<1|\lambda -\lambda_0| \cdot \|R(\lambda )\|<1, one has:

R(λ)=∑n=0∞(λ0−λ)nR(λ0)n+1 R(\lambda ) =\sum_{n=0}^{\infty}(\lambda_0-\lambda )^nR(\lambda_0)^{n+1}

(b) For all λ∈Δ\lambda \in \Delta and every integer n⩾0n\geqslant 0, one has

∂n∂λnR(λ)=(−1)nn\frac{\partial^n}{\partial \lambda^n}R(\lambda ) = (-1)^nn!R(λ\lambda )n+1^{n+1}.

(c) If R is holomorphic at infinity, then there exist z,j,x∈Az, j, x\in A such that z2=0z^2= 0, j2=j,zj=jz=0,x∈jAjj^2=j,zj=jz= 0,x\in jAj, and R(λ)=z+jR(λ,x)R(\lambda ) =z+jR(\lambda , x) for ∣λ∣|\lambda | sufficiently large. (Write R(λ)=∑n=0∞cnλ−nR(\lambda ) =\sum^{\infty}_{n=0}c_n\lambda^{-n} for ∣λ∣>λ0|\lambda |> \lambda_0, and express that R satisfies (5). One may put c0=z,c1=j,c2=x.c_0=z,c_1=j,c_2=x.)

(d) In order that there exist a∈Aa\in A such that R is the restriction to Δ\Delta of the resolvent λ↦R(a,λ)\lambda \mapsto R(a, \lambda ) of aa, it is necessary and sufficient that there exist λ0∈Δ\lambda_0\in \Delta such that R(λ0)R(\lambda_0) is invertible.

(e) For every x∈Ax\in A, the function λ↦∑n=0∞(λ0−λ)nxn+1\lambda \mapsto \sum^{\infty}_{n=0}(\lambda_0-\lambda )^nx^{n+1} satisfies (5) for ∣λ−λ0∣⋅∥x∥<1|\lambda -\lambda_0| \cdot \|x\|<1.

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