What does it mean that the "resolvent mapping is analytical"?

Question

This is taken directly from our lecture notes, where $H$ is a complex Hilbert space.

Defintion. Let $T: H \supset \text{dom}(T) \subset H$ be densely defined.

1. The set $\varrho(T) := \{\lambda \in \mathbb{C}: \lambda I - T: \text{dom}(T) \to H$ has a bounded inverse in $H$} is the resolvent set of $T$.

2. The mapping $R: \varrho(T) \to L(H)$, $R_{\lambda} = R(\lambda) := (\lambda I - T)^{-1}$ is called the resolvent mapping.

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Theorem. Let $T$ be as above.

1. $\varrho(T)$ is open.
2. The resolvent mapping is analytic and the resolvent identity $$ R_{\lambda} - R_{\mu} = (\mu - \lambda) R_{\lambda} R_{\mu} $$ holds.

Proof. 1. Let $\lambda_0 \in \varrho(T)$ and $| \lambda - \lambda_0 | < \| R_{\lambda_0} \|^{-1}$. Then $$ \tag{1} \lambda I - T = (\lambda_0 I - T) + (\lambda - \lambda_0) I = (\lambda_0 I - T) \underbrace{\left[ I - (\lambda - \lambda_0)(\lambda_0 I - T)^{-1}\right]}_{[\ldots]^{-1} = \sum_{n = 0}^{\infty} ((\lambda - \lambda_0)(\lambda_0 I - T)^{-1})^n}. $$ Hence $\lambda I - T$ is invertible.

2. A formal calculations yields $$ (\lambda I - T)(\mu I - T) \left[ ( \lambda I - T)^{-1} - (\mu I - T)^{-1}\right] = (\mu - \lambda) I $$ and a similar identity can be derived by multiplying from the right side. So, only the (easy) inspection of domains is missing. $\square$

Questions

1. What does it mean for the resolvent mapping to be analytic? Is it that we can write it as $R_{\lambda} = \sum_{n = 0}^{\infty} a_n T^n$? Does this follow from $(1)$ since $(1)$ implies $$ (\lambda I - T)^{-1} = \sum_{n = 0}^{\infty} ((\lambda_0 - \lambda) (\lambda_0 I - T))^n \cdot (\lambda_0 I - T)^{-1} = \sum_{n = 0}^{\infty} (\lambda_0 - \lambda)^n (\lambda_0 I - T)^{n - 1} \ ? $$

   2. The wording at the beginning of the proof of 1. troubles me a little. We assume that $\varrho(T) \ne \emptyset$, right? So would the exact wording of the beginning be:

Let $\lambda_0 \in \rho(T)$. Then there exists a $\lambda \in \mathbb C$ such that $| \lambda - \lambda_0 | < \ldots$?

Answer 1

To your first question: You can define holomorphy for Banach-space-valued functions in a pretty straight forward fashion. Let $D \subseteq \mathbb C$ be a non-empty, open set, $E$ some Banach space and $f: D \to E$ some function. Then $f$ is called holomorphic if the limit $$ f'(z) := \lim_{z \to z_0} \frac{1}{z - z_0}(f(z) - f(z_0)) \in E$$ exists for each $z_0 \in D$. Now it shouldn't suprise that most of the results that you know from complex analysis still hold in this Banach space setting. For example, Cauchy's theorem holds, locally uniform limits of holomorphic functions are again holomorphic, each holomorphic function has a power series representation and you can even do Laurent series expansion for isolated points.

Now you might notice that you can write the resolvent of an operator as power series by means of the Neumann series, i.e., $$ R(\lambda, T) = \sum_{n = 0}^\infty (\lambda - \mu)^n (-1)^n R(\mu, T)^{n + 1} \qquad \text{for} \quad \lvert \lambda - \mu \rvert < \lVert R(\mu, T) \rVert^{-1}.$$ This is a convergent power series in the Banach space $\mathcal L(E)$ of bounded linear operator on $E$. Hence the resolvent is a holomorphic function in the sense we discussed above and therefore analytical. Here we call a function $f: D \to E$ analytical, if for each $z_0 \in D$ it can be represeted locally by a power series $$f(z) = \sum_{n = 0}^\infty (z - z_0)^n x_n$$ for $x_n \in E$. So by choosing $x_n := (-1)^n R(\mu, T)^{n + 1} \in \mathcal L(E)$, we see that the resolvent map is analytical.

The answer to your second question is more mundane: If the resolvent set is empty, it is trivially open. So as there is nothing to show in this case, the author just supposes that it is not empty.

I hope I could help you out a bit :)