Neumann series

Remarks

Proof idea: $\norm{T^n} \le \norm{T}^n$, so the series is dominated by a geometric series and converges absolutely in the Banach space of bounded operators on $X$. See absolutely convergent series in a Banach space converge. 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 $k \ge 1$.

Going deeper

For a student

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.