# Invertibility of Identity Minus Operator

## Theorem

Suppose $B$ is a Banach space and $T \in \mathfrak{L}(B, B)$, the space of bounded linear operators on $B$. If $\| T \|_* < 1$ in the operator norm $\|\cdot\|_*$:, then $I - T$ is invertible and has inverse

$$(I - T)^{-1} = I + T + T^2 + T^3 + \ldots.$$

## Proof

Define $S_n = I + T + T^2 + \ldots + T^n$. We first argue that $S_n$ converges to a bounded linear operator $S\in \mathfrak{L}(B,B)$. For any $n > m$,

 $\ds \norm{S_n - S_m}_*$ $=$ $\ds \norm{T^{m+1} + \ldots + T^n}_*$ $\ds$ $\leq$ $\ds \norm{T^{m+1} }_* + \ldots + \norm{T^n}_*$ by Triangle Inequality and Operator Norm is Norm $\ds$ $\leq$ $\ds \norm{T}_*^{m+1} + \ldots + \norm{T}_*^n$ by repeated use of Operator Norm on Banach Space is Submultiplicative on each term $\ds$ $\leq$ $\ds \sum_{j=m+1}^\infty \norm{T}_*^{j}$ since each term is positive.

Since $\|T\|_* < 1$ by assumption, this is a tail of a converging infinite series; therefore, by Tail of Convergent Series tends to Zero, this can be made arbitrarily small. Hence the sequence $S_n$ is in fact Cauchy; by Set of Bounded Linear Operators on Banach Space is Banach Space, it converges in the operator norm to some bounded linear $S = \lim_{n\to\infty} S_n$.

Next, expand $(I-T)S_k = (I-T)(I + T + \ldots + T^k)$ to yield $I - T^{k+1}$. We argue that $I - T^{k+1} \to I$ in norm:

 $\ds \norm{ (I - T^{k+1}) - I }_*$ $=$ $\ds \norm{ - T^{k+1} }_*$ $\ds$ $=$ $\ds \norm { T^{k+1} }_*$ by Operator Norm is Norm $\ds$ $\leq$ $\ds \norm { T }^{k+1}$ by repeated application of Operator Norm on Banach Space is Submultiplicative $\ds$ $\to$ $\ds 0$ since $\norm{T}_* < 1$ by assumption.

Taking limits on the left-hand side, we get $\lim_{k\to\infty} (I-T)S_k = (I-T)S = I$.

A similar computation shows us that $\lim_{k\to\infty} S_k(I-T) = S(I-T) = I$. So $S=I+T+T^2+\ldots$ is the bounded two-sided inverse of $I-T$.

$\blacksquare$