TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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Resolvent is Banach-valued holomorphic

Statement

Let $A$ be a unital complex Banach algebra and let $a \in A$ with resolvent set $ρ_{A} (a)$ and resolvent $R (z, a) = (z 1 - a)^{- 1}$ (Spectrum and resolvent set in a Banach algebra). Then:

$ρ_{A} (a)$ is an open subset of $C$ , so $σ_{A} (a)$ is closed;
for every $z_{0} \in ρ_{A} (a)$ and every $h \in C$ with $∣ h ∣ ∥ R (z_{0}, a) ∥ < 1$ the Neumann expansion $R (z_{0} + h, a) = \sum_{n \geq 0} (- h)^{n} R (z_{0}, a)^{n + 1}$ converges in $A$ and exhibits $z_{0} + h \in ρ_{A} (a)$ ;
the map $z \mapsto R (z, a)$ is holomorphic on $ρ_{A} (a)$ in the norm sense, with derivative $R^{'} (z, a) = - R (z, a)^{2} (z \in ρ_{A} (a)),$ and in particular it is norm continuous there, with the local estimate $∥ R (z_{0} + h, a) - R (z_{0}, a) ∥ \leq ∣ h ∣ ∥ R (z_{0}, a) ∥^{2} / (1 - ∣ h ∣ ∥ R (z_{0}, a) ∥)$ for $∣ h ∣ ∥ R (z_{0}, a) ∥ < 1$ .

Facts & Assumptions

Given: A unital complex Banach algebra $A$ , an element $a \in A$ , a point $z_{0} \in ρ_{A} (a)$ with $R_{0} : = R (z_{0}, a)$ , and $h \in C$ with $∣ h ∣ ∥ R_{0} ∥ < 1$ .

[L1]

$A$ is complete, the norm is submultiplicative with $∥ 1 ∥ = 1$ , and multiplication is associative and bilinear (Unital Banach algebra).

[L2]

$R (z, a)$ is the unique two-sided inverse of $z 1 - a$ , so $(z 1 - a) R (z, a) = R (z, a) (z 1 - a) = 1$ , and $R (z_{0} + h, a)$ exists exactly when $(z_{0} + h) 1 - a$ is invertible (Spectrum and resolvent set in a Banach algebra).

[L3]

If $∥ y ∥ < 1$ then $1 - y$ is invertible with $(1 - y)^{- 1} = \sum_{n \geq 0} y^{n}$ and $∥ (1 - y)^{- 1} ∥ \leq 1 / (1 - ∥ y ∥)$ (Neumann series).

[L4]

Inversion is continuous on the invertible group, and $ρ_{A} (a)$ is therefore open: it is the preimage of the open set $A^{\times}$ under the continuous map $z \mapsto z 1 - a$ (Invertible group is open and inversion is continuous).

Proof

technique · direct

1.1L1L2algebra

Put $x : = - h R_{0}$ , so that $∥ x ∥ \leq ∣ h ∣ ∥ R_{0} ∥ < 1$ by [L1], and $(z_{0} + h) 1 - a = (z_{0} - a) (1 - x)$ : indeed $(z_{0} - a) (1 + h R_{0}) = (z_{0} - a) + h (z_{0} - a) R_{0} = (z_{0} - a) + h$ , using $(z_{0} - a) R_{0} = 1$ from [L2].

2.1step 1.1L3

By [L3] applied to $x$ , the element $1 - x$ is invertible with $(1 - x)^{- 1} = \sum_{n \geq 0} (- h)^{n} R_{0}^{n}$ and $∥ (1 - x)^{- 1} ∥ \leq 1 / (1 - ∣ h ∣ ∥ R_{0} ∥)$ .

3.1step 1.1step 2.1L2L4

By [step 1.1] and [step 2.1], $(z_{0} + h) 1 - a$ is a product of two invertible elements, hence invertible, with $R (z_{0} + h, a) = (1 - x)^{- 1} R_{0} = \sum_{n \geq 0} (- h)^{n} R_{0}^{n + 1}$ ; combined with [L4] this shows that $ρ_{A} (a)$ is open, which is claim 1.

4.1step 3.1L4L1

The map $z \mapsto R (z, a)$ is norm continuous at $z_{0}$ : from [step 3.1], $R (z_{0} + h, a) - R_{0} = \sum_{n \geq 1} (- h)^{n} R_{0}^{n + 1}$ , whose norm is at most $\sum_{n \geq 1} ∣ h ∣^{n} ∥ R_{0} ∥^{n + 1} = ∣ h ∣ ∥ R_{0} ∥^{2} / (1 - ∣ h ∣ ∥ R_{0} ∥)$ , and this tends to $0$ with $h$ ; independently, continuity of inversion [L4] applied to the continuous map $z \mapsto z 1 - a$ gives the same conclusion.

4.2step 3.1L1algebra

For nonzero $h$ with $∣ h ∣ ∥ R_{0} ∥ < 1$ , divide the expansion of [step 3.1] by $h$ after subtracting $R_{0}$ : $\frac{R (z_{0} + h, a) - R_{0}}{h} + R_{0}^{2} = \sum_{n \geq 2} (- 1)^{n} h^{n - 1} R_{0}^{n + 1}$ . The right-hand side converges in $A$ and has norm at most $\sum_{n \geq 2} ∣ h ∣^{n - 1} ∥ R_{0} ∥^{n + 1} = \frac{∣ h ∣ ∥ R_{0} ∥^{3}}{1 - ∣ h ∣ ∥ R_{0} ∥}$ , which tends to $0$ as $h \to 0$ .

5.1step 4.2L4

Thus the norm difference quotient of $h \mapsto R (z_{0} + h, a)$ at $0$ converges to $- R_{0}^{2}$ . Since $z_{0}$ was arbitrary in the open set $ρ_{A} (a)$ , the resolvent is Banach-valued holomorphic there and $R^{'} (z_{0}, a) = - R (z_{0}, a)^{2}$ .

6.1step 3.1step 4.1step 5.1∎

The three claims are established: claim 1 by [step 3.1], claim 3 together with its continuity and estimate by [step 4.1] and [step 5.1], and claim 2 is exactly the expansion of [step 3.1].

Resolvent is Banach-valued holomorphic

Statement

Facts & Assumptions

Proof

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