Neumann Series Invertibility

Question

It is a fairly quick exercise to show that for continuous operators $T$ on a Banach space $X$ with $\|T\| < 1$ we have the geometric series (aka Neumann series) type identity: $$ (I - T)^{-1} = \sum_{k = 0}^{\infty} T^k $$ Showing this also shows the operator $(I-T)$ is invertible. My question is, is there a general way to prove this without a series approach? I attempted to prove that $I-T$ was a bijection. Injectivity comes from the fact that: $$ (I-T)x = 0 \iff x = Tx \iff x = 0 $$ Witht he last implication following from the fact that $\|T\| < 1$. However, I am not sure how to prove surjectivity (if there is a way).

Answer 1

As you can use your argument with $T^*$, you automatically get that $$ \overline{\text{ran}(I-T)}=\ker(I-T^*)^\perp=H. $$ So $I-T$ has dense range.

Now let $z\in H$. By the above, there exists $\{x_n\}\subset X$ with $x_n-Tx_n\to z$. Now fix $\varepsilon>0$. Then there exists $n_0$ such that whenever $n\geq n_0$, $\|x_n-Tx_n-z\|<(1-\|T\|)\varepsilon/2$. Then, for any $m,n\geq n_0$, \begin{align} \|x_n-x_m\|&=\|x_n-Tx_n-z+Tx_n-Tx_m+z-(x_m-Tx_m)\| \\ \ \\ &\leq\|x_n-Tx_n-z\|+\|Tx_n-Tx_m\|+\|x_m-Tx_m-z\|\\ \ \\ &\leq 2(1-\|T\|)\varepsilon/2+\|T\|\,\|x_n-x_m\|. \end{align} We may rewrite the above as $$ \|x_n-x_m\|<\varepsilon. $$ That is, we have shown that $\{x_n\}$ is Cauchy. As $X$ is Banach, there exists $x\in X$ with $x_n\to x$. As $T$ is bounded, $Tx_n\to Tx$. Then $z=\lim_n x_n-Tx_n=x-Tx$. So $I-T$ is surjective.