---
tag: 8QJ91709V2
kind: statement
role: theorem
name: Neumann series
msc: [47A05, 46B99]
uses: [06GZ956657, 03TM135G0N, 06EKW1AKXF, CPAD5W1CGW, 082MGNWNA8, 01X5CVV1Q5]
about: [CPAD5W1CGW]
sources:
  - url: https://en.wikipedia.org/wiki/Neumann_series
---

## Statement

Let $X$ be a [[Banach space|06GZ956657]], and let $T \colon X \to X$ be a [[continuous linear map|03TM135G0N]] with [[operator norm|06EKW1AKXF]] $\norm{T} < 1$.
Write $I$ for the identity map of $X$ and $T^0 = I$.
Then the following hold.

1. The series $\sum_{n=0}^{\infty} T^n$ converges in operator norm.
2. The operator $I - T$ is [[invertible|CPAD5W1CGW]], and
$$(I - T)^{-1} = \sum_{n=0}^{\infty} T^n.$$
3. $\norm{(I - T)^{-1}} \le \dfrac{1}{1 - \norm{T}}$.

## Remarks

Proof idea: $\norm{T^n} \le \norm{T}^n$, so the series is dominated by a [[geometric series|01X5CVV1Q5]] and converges absolutely in the Banach space of bounded operators on $X$.
See [[absolutely convergent series in a Banach space converge|082MGNWNA8]].
The partial sums $S_N = \sum_{n=0}^{N} T^n$ satisfy $(I - T) S_N = S_N (I - T) = I - T^{N+1}$, and $T^{N+1} \to 0$.
Letting $N \to \infty$ gives both inverse identities, and the norm bound follows from the geometric series.

Example: on $\ell^2$, let $T(x_1, x_2, \ldots) = (0, \tfrac12 x_1, \tfrac12 x_2, \ldots)$, half the right shift.
Then $\norm{T} = \tfrac12$, so $I - T$ is invertible with $\norm{(I - T)^{-1}} \le 2$.

Completeness is needed.
On the space of finitely supported sequences with the $\ell^2$ norm, $T$ as above has norm $\tfrac12$.
But the solution of $(I - T)x = (1, 0, 0, \ldots)$ is $x = (1, \tfrac12, \tfrac14, \ldots)$, which is not finitely supported, so $I - T$ is not surjective there.

A consequence: if $A$ is invertible and $\norm{B - A} < 1 / \norm{A^{-1}}$, then $B$ is invertible, so the invertible operators form an open set.
The series also converges, with the same sum, whenever $\norm{T^k} < 1$ for some $k \ge 1$.

## Explanations

### L1 human

For numbers, $1/(1 - t) = 1 + t + t^2 + \cdots$ when $\abs{t} < 1$.
The Neumann series says the same formula works for operators, with the absolute value replaced by the operator norm.
So an operator that is a small perturbation of the identity can always be undone, and the inverse is computed by a convergent series.

### L2 human

The Neumann series is the main tool to prove that invertibility is stable under small perturbations.
It shows that the set of invertible operators is open and that inversion is continuous, indeed analytic, on it.
In spectral theory it shows that the spectrum of $T$ lies in the closed disc of radius $\norm{T}$, and that the resolvent is analytic.
For integral equations $x - Kx = y$ with a small kernel $K$, the series gives the solution by iteration.
The series is named after Carl Neumann, who used it in potential theory.