Operator inversion from summable bounds on powers

Statement

Let $T$ be a bounded operator on a Banach space. If $\sum_{k=0}^\infty\|T^k\|<\infty$ , then

(I+T)^{-1}=\sum_{k=0}^\infty(-T)^k

in operator norm. Multiplication of a finite partial sum by $I+T$ , on either side, gives $I+(-1)^NT^{N+1}$ ; its remainder tends to zero. Thus the inverse is two-sided. This criterion does not require $\|T\|<1$ .

Factorial gains from increasing degree

Suppose nested subspaces $X_b$ satisfy $X_0=X$ , $T(X_b)\subseteq X_{b+1}$ , and $\|Tx\|\le K\|x\|/[(b+1)(b+2)]$ for $x\in X_b$ . Applying this bound successively yields

\|T^k\|\le\frac{K^k}{k!(k+1)!},

which is summable for every finite $K$ . Vanishing of low-degree coefficients provides such a filtration for suitable radial integral operators.