Question on Hilbert Space: $\;z \in ρ(T) \rightarrow (z-T)^{-1} \in \mathcal B(H)\;$ is Analytic

Question

Let $\;T:H \rightarrow H\;$ be a bounded operator. Show that the function $\;z \in ρ(T) \rightarrow (z-T)^{-1} \in \mathcal B(H)\;$ is analytic, where $\;ρ(T)\;$ is the resolvent set : $\;ρ(T)=\{ z\in \mathbb C: (z-T)^{-1} \in \mathcal B(H)\}\;$

Here is the proof, the professor gave us on class:

By Neumann's Theorem for $\; \vert z-z_0 \vert \lt \frac{1}{\vert \vert (z_0 -T)^{-1} \vert \vert}\;$ we get \begin{align} (z-T)^{-1}& = (z-z_0+z_0-T)^{-1}\\ &=(z_0 -T)^{-1}\left[I+(z-z_0)(z_0-T)^{-1}\right]^{-1}\\ &=(z_0 -T)^{-1}\sum_{n=0}^{\infty}(-1)^{n}(z-z_0)^{n}\left[(z_0 -T)^{-1}\right]^{n}\\ &=\sum_{n=0}^{\infty} (-1)^{n} (z-z_0)^{n} \left[(z_0 -T)^{-1}\right]^{n+1} \end{align} Since the operator $\;(z-T)^{-1}\;$ can be written (locally) as a convergent power series, the function is analytic on $\;ρ(T)\;$.

Neumann's Theorem below:

My questions:

1. How do I know that series is convergent? Is it because for $\; \vert z-z_0 \vert \lt \frac{1}{\vert \vert (z_0 -T)^{-1} \vert \vert}\;$ one can show $\;z \in ρ(T)\;$? However it's not very clear to me...
2. Why did he use the term "locally"?

I would appreciate any help! Thanks in advance!

Answer 1

$1)$ We know the series is convergent because of the lemma. Since $\|(z-z_0)(z_0-T)^{-1}\|<1$, the lemma implies that $I+(z-z_0)(z_0-T)^{-1}$ is invertible and the series representation is convergent. The convergence of the series is also consequence of the following:

If $X$ is a Banach space, $\{x_n\}$ a sequence in $X$, and $\sum_n\|x_n\|<\infty$, then $\sum_nx_n$ is convergent, and $\|\sum_nx_n\|\leq\sum_n\|x_n\|$.

$2)$ The term "locally" is as an antonym for "globally". There are $z\in\mathbb{C}$ such that $z-T$ is invertible, but whose inverse cannot be represented as a power series in $z-z_0$ (similarly to holomorphic functions).