Neumann series and small perturbations of bounded inverses

For every $n$ one has $\mathrm{\parallel }{R}^n\mathrm{\parallel }\le \mathrm{\parallel }R{\mathrm{\parallel }}^n$, and ${\sum }_n\mathrm{\parallel }R{\mathrm{\parallel }}^n$ converges to $1\mathrm{/}(1-\mathrm{\parallel }R\mathrm{\parallel })$.

The space $\mathcal{B}(X)$ is Banach, so the absolutely convergent series ${\sum }_n{R}^n$ converges in operator norm to some $S\in \mathcal{B}(X)$. For every $N$, the finite triangle inequality and [step 1.1] give $\mathrm{\parallel }{S}_N\mathrm{\parallel }\le {\sum }_{n<N}\mathrm{\parallel }{R}^n\mathrm{\parallel }\le {\sum }_{n=0}^{\mathrm{\infty }}\mathrm{\parallel }R{\mathrm{\parallel }}^n=\frac{1}{1-\mathrm{\parallel }R\mathrm{\parallel }}\mathrm{.}$ Since $S_N\to S$ and the norm is continuous, taking the limit yields $\mathrm{\parallel }S\mathrm{\parallel }\le 1\mathrm{/}(1-\mathrm{\parallel }R\mathrm{\parallel })$.

For every $N$ one has $(I-R)S_N=S_N(I-R)=I-R^N$, and $\mathrm{\parallel }{R}^N\mathrm{\parallel }\le \mathrm{\parallel }R{\mathrm{\parallel }}^N\to 0$, so $R^N\to 0$.

From [step 2.1] and [step 2.2], $(I-R)S={\lim }_N(I-R)S_N={\lim }_N(I-R^N)=I$ and likewise $S(I-R)=I$: the first limit holds because $\mathrm{\parallel }(I-R)(S-S_N)\mathrm{\parallel }\le (1+\mathrm{\parallel }R\mathrm{\parallel })\mathrm{\parallel }S-S_N\mathrm{\parallel }\to 0$.

Hence $I-R$ is invertible with inverse $S={\sum }_{n\ge 0}{R}^n$ and $\mathrm{\parallel }(I-R{)}^{-1}\mathrm{\parallel }\le 1\mathrm{/}(1-\mathrm{\parallel }R\mathrm{\parallel })$, which is claim 1.

For the perturbation, $A+E=A(I+A^{-1}E)$ and $\mathrm{\parallel }-A^{-1}E\mathrm{\parallel }=\mathrm{\parallel }{A}^{-1}E\mathrm{\parallel }<1$, so by [step 3.1] applied to $R:=-A^{-1}E\in \mathcal{B}(X)$ the operator $I+A^{-1}E$ is invertible with bounded inverse, and therefore $(A+E{)}^{-1}=(I+A^{-1}E{)}^{-1}{A}^{-1}\in \mathcal{B}(Y,X)$, which is claim 2.

Claims 1 and 2 are exactly the two parts of the statement.