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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 aA with a<1, and for NN write SN:=n=0Nan, a finite sum with a0:=1 (Unital Banach algebra). Then

  1. the series n0an converges in A, and its sum S:=limNSN satisfies (1a)S=S(1a)=1; in particular 1a is invertible with (1a)1=S (Invertible element and general linear group of a Banach algebra);
  2. for every NN the tail estimate (1a)1SN    aN+11a holds. The hypothesis a<1 is not symmetric: the estimate is in terms of a, and 1a is invertible whenever a lies in the open unit ball.

Facts & Assumptions

Given: A unital complex Banach algebra A, an element aA with a<1, the partial sums SN=n=0Nan, and the number q:=a[0,1).

[L1]

A is complete under its norm, 1=1, and xyxy for all x,yA; multiplication is associative and bilinear (Unital Banach algebra).

[L2]

An element cA is invertible exactly when there is bA with cb=bc=1, and that b is then unique, written c1 (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
1.1

For every nN one has anan=qn: this holds at n=0 because a0=1=1=q0, and inductively an+1=anaanaqnq=qn+1.

L1algebra
1.2

For every NN the telescoping identities (1a)SN=1aN+1 and SN(1a)=1aN+1 hold, by distributivity and anan+1=an(1a) summed over 0nN.

L1algebra
1.3

Multiplication is jointly continuous in the norm: for x,x,y,yA one has xyxyxyy+xxy by [L1], so xx and yy force xyxy.

L1algebra
2.1

Since q[0,1), the geometric series satisfies n0qn=1/(1q) and its tails satisfy n>Nqn=qN+1/(1q); with [step 1.1] this gives n0an1/(1q)<.

step 1.1algebra
3.1

The series n0an is absolutely convergent, so it converges to an element S=limNSNA by [L3] and completeness of A.

L3step 2.1L1
4.1

Since aN+10 in A by [step 1.1] and qN+10, letting N in the identities of [step 1.2] is legitimate: (1a)SN(1a)S and SN(1a)S(1a) by [step 1.3] and [step 3.1], while the right hand sides 1aN+1 tend to 1; hence (1a)S=1 and S(1a)=1.

step 1.2step 1.3step 3.1step 1.1
5.1

By [L2] the element 1a is invertible with (1a)1=S=n0an, which proves claim 1; moreover for every N the difference of the sum and the partial sum is the tail (1a)1SN=n>Nan, whose norm is at most n>Nqn=qN+1/(1q) by [step 1.1] and [step 2.1], which is claim 2.

step 4.1step 2.1L2algebra

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