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

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Neumann series

Statement

Let $A$ be a unital complex Banach algebra, let $a \in A$ with $∥ a ∥ < 1$ , and for $N \in N$ write $S_{N} : = \sum_{n = 0}^{N} a^{n}$ , a finite sum with $a^{0} : = 1$ (Unital Banach algebra). Then

the series $\sum_{n \geq 0} a^{n}$ converges in $A$ , and its sum $S : = \lim_{N \to \infty} S_{N}$ satisfies $(1 - a) S = S (1 - a) = 1$ ; in particular $1 - a$ is invertible with $(1 - a)^{- 1} = S$ (Invertible element and general linear group of a Banach algebra);
for every $N \in N$ the tail estimate $∥ (1 - a)^{- 1} - S_{N} ∥ \leq \frac{∥ a ∥^{N + 1}}{1 - ∥ a ∥}$ holds. The hypothesis $∥ a ∥ < 1$ is not symmetric: the estimate is in terms of $∥ a ∥$ , and $1 - a$ is invertible whenever $a$ lies in the open unit ball.

Facts & Assumptions

Given: A unital complex Banach algebra $A$ , an element $a \in A$ with $∥ a ∥ < 1$ , the partial sums $S_{N} = \sum_{n = 0}^{N} a^{n}$ , and the number $q : = ∥ a ∥ \in [0, 1)$ .

[L1]

$A$ is complete under its norm, $∥ 1 ∥ = 1$ , and $∥ x y ∥ \leq ∥ x ∥ ∥ y ∥$ for all $x, y \in A$ ; multiplication is associative and bilinear (Unital Banach algebra).

[L2]

An element $c \in A$ is invertible exactly when there is $b \in A$ with $c b = b c = 1$ , and that $b$ is then unique, written $c^{- 1}$ (Invertible element and general linear group of a Banach algebra).

[L3]

A normed space $V$ is a Banach space if and only if every absolutely convergent series in $V$ converges (Series criterion for Banach spaces).

Proof

technique · direct

1.1L1algebra

For every $n \in N$ one has $∥ a^{n} ∥ \leq ∥ a ∥^{n} = q^{n}$ : this holds at $n = 0$ because $∥ a^{0} ∥ = ∥ 1 ∥ = 1 = q^{0}$ , and inductively $∥ a^{n + 1} ∥ = ∥ a^{n} a ∥ \leq ∥ a^{n} ∥ ∥ a ∥ \leq q^{n} q = q^{n + 1}$ .

1.2L1algebra

For every $N \in N$ the telescoping identities $(1 - a) S_{N} = 1 - a^{N + 1}$ and $S_{N} (1 - a) = 1 - a^{N + 1}$ hold, by distributivity and $a^{n} - a^{n + 1} = a^{n} (1 - a)$ summed over $0 \leq n \leq N$ .

1.3L1algebra

Multiplication is jointly continuous in the norm: for $x, x^{'}, y, y^{'} \in A$ one has $∥ x y - x^{'} y^{'} ∥ \leq ∥ x ∥ ∥ y - y^{'} ∥ + ∥ x - x^{'} ∥ ∥ y^{'} ∥$ by [L1], so $x^{'} \to x$ and $y^{'} \to y$ force $x^{'} y^{'} \to x y$ .

2.1step 1.1algebra

Since $q \in [0, 1)$ , the geometric series satisfies $\sum_{n \geq 0} q^{n} = 1 / (1 - q)$ and its tails satisfy $\sum_{n > N} q^{n} = q^{N + 1} / (1 - q)$ ; with [step 1.1] this gives $\sum_{n \geq 0} ∥ a^{n} ∥ \leq 1 / (1 - q) < \infty$ .

3.1L3step 2.1L1

The series $\sum_{n \geq 0} a^{n}$ is absolutely convergent, so it converges to an element $S = \lim_{N} S_{N} \in A$ by [L3] and completeness of $A$ .

4.1step 1.2step 1.3step 3.1step 1.1

Since $a^{N + 1} \to 0$ in $A$ by [step 1.1] and $q^{N + 1} \to 0$ , letting $N \to \infty$ in the identities of [step 1.2] is legitimate: $(1 - a) S_{N} \to (1 - a) S$ and $S_{N} (1 - a) \to S (1 - a)$ by [step 1.3] and [step 3.1], while the right hand sides $1 - a^{N + 1}$ tend to $1$ ; hence $(1 - a) S = 1$ and $S (1 - a) = 1$ .

5.1step 4.1step 2.1L2algebra∎

By [L2] the element $1 - a$ is invertible with $(1 - a)^{- 1} = S = \sum_{n \geq 0} a^{n}$ , which proves claim 1; moreover for every $N$ the difference of the sum and the partial sum is the tail $(1 - a)^{- 1} - S_{N} = \sum_{n > N} a^{n}$ , whose norm is at most $\sum_{n > N} q^{n} = q^{N + 1} / (1 - q)$ by [step 1.1] and [step 2.1], which is claim 2.

Neumann series

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources