Weaker Assumption for Convergence of the Neumann series

Question

I learned about a theorem which states:

Let $A \in \mathbb C^{n \times n}$. The Neumann series converges if and only if it holds for the spectral radius that $r(A) < 1$. In this case, $I − A$ is invertible and we have $$ \sum_{n = 0}^\infty A^n = (I - A)^{-1}.$$

I could show one implication of this theorem in a Banach space setting:

Let $X$ a Banach space and $T: X \to X$ a compact operator. If the Neumann series convergences it holds that $r(T) < 1$.

But I have no idea how to show the other implication in this setting.

I wondered if one can make such a theorem work for operators on Banach spaces or Hilbert spaces? Is there a analogous theorem for a suitable infinite dimensional setting (maybe compact operators)? If there is one, I would really enjoy some literature advice, because I couldn't find anything related to that using google. Thanks in advance :)

Do you remember a formula for $r(T)$ that uses $\lVert T^n\rVert$? — Daniel Fischer, Dec 30 '16 at 20:41
I have added a proof and commented about your last paragraph here https://math.stackexchange.com/questions/242425/neumann-series-and-spectral-radius — R. W. Prado, Aug 06 '21 at 03:40

score 7 · Accepted Answer · answered Dec 30 '16 at 20:46

Recall that the spectral radius satisfies

$$r(T) = \lim_{n\to \infty}\, \lVert T^n\rVert^{1/n}.$$

Then choose any $q \in \bigl(r(T),1\bigr)$. For all large enough $n$, we have $\lVert T^n\rVert < q^n$, and we see that the Neumann series is (absolutely) convergent when $r(T) < 1$.

Weaker Assumption for Convergence of the Neumann series

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