Neumann series of resolvent operator

Question

Let $X$ be a Banach space on $\mathbb{C}$ and $T\in \mathcal{B}(X)$. We know that the functional $\lambda I - T$ is a bijection and has a bounded inverse $R_\lambda(T)$ for all $\lambda \in \rho(T)\subseteq \mathbb{C}$. Furthermore, $\rho(T)\subseteq \mathbb{C}\setminus B_{\|T\|} $ and there $$R_\lambda(T)= \sum_{n=0}^\infty \frac{T^n}{\lambda^{n+1}}.$$

If we define the spectral radius $r(T)$ of $T$ as $$ r(T)=\sup_{\lambda \in \mathbb{C}\setminus\rho (T)}|\lambda|$$ Reed-Simon say that the series expansion written above for $R_\lambda(T)$ is valid also for $\lambda\in B_{r(T)}$, furthermore $r(T)\le \|T\|$.

I want to prove it using a topic that involves the weak analyticity of $R_\lambda(T)$ (to avoid to develop a theory about analytic functions with values in an operator space).

So, I think in that way. I know that $R_\lambda(T)$ satisfies the following properties:

$$ \rho(T) \ni \lambda \mapsto \phi(R_\lambda(T)) \in \mathbb{C} \text{ is analytic } \forall \phi \in \mathcal{B}(X)^* $$

So the function $f(z)=\phi(R_{\frac{1}{z}}(T))$ has the series $$ \sum_{n=0}^\infty z^{n+1}\phi(T^n) $$

as power series centered in $0$ for $|z|<\|T\|^{-1}$. But that means that the same expansion is valid for $|z|<r(T)^{-1}$, for the coefficient of the series expansion in $z=0$ only depend on the value at $0$.

Can someone please help me to conclude…

Answer 1

Let us assume that for each $\phi \in \mathcal{B}(X)^{*}$ and each $z \in \mathbb{C}$ with $0 < |z| < r(T)^{-1}$ we have the identity \begin{equation} \phi (R_{\tfrac{1}{z}}(T)) = \sum_{k=0}^{\infty} z^{k+1} \phi (T^{k}) . \tag{1} \end{equation} If we knew for each $z\in \mathbb{C}$ with $0 < |z| < r(T)^{-1}$ that the series $(\sum_{k=0}^{n} z^{k+1} T^{k} )_{n\in\mathbb{N}}$ converges in the norm topology, we can use continuity and $(1)$ to deduce that \begin{equation} \phi (R_{\tfrac{1}{z}}(T)) = \phi \Big( \sum_{k=0}^{\infty} z^{k+1} T^{k} \Big) \end{equation} for all $\phi \in \mathcal{B}(X)^{*}$ and $z\in \mathbb{C}$ with $0 < |z| < r(T)^{-1}$. It then follows from a Hahn-Banach theorem that \begin{equation} R_{\tfrac{1}{z}}(T) = \sum_{k=0}^{\infty} z^{k+1} T^{k} \end{equation} for all $z\in \mathbb{C}$ with $0 < |z| < r(T)^{-1}$, which is equivalent to the condition that \begin{equation} R_{\lambda }(T) = \sum_{k=0}^{\infty} \lambda^{-(k+1)} T^{k} \end{equation} for all $\lambda \in \mathbb{C}$ with $|\lambda | > r(T)$. So our problem has been reduced to showing that for each $\phi \in \mathcal{B}(X)^{*}$ and $z\in \mathbb{C}$ with $0 < |z| < r(T)^{-1}$, the series $(\sum_{k=0}^{n} z^{k+1} T^{k} )_{n\in\mathbb{N}}$ converges in the norm topology.

Let $s\in (0, \infty )$ with $s < r(T)^{-1}$. We will show that for every $z\in \mathbb{C}$ with $0 < |z| < s$ the series $(\sum_{k=0}^{n} z^{k+1} T^{k} )_{n\in\mathbb{N}}$ converges absolutely. Let $\phi \in \mathcal{B}(X)^{*}$. By $(1)$ the series $(\sum_{k=0}^{n} s^{k+1} \phi (T^{k}))_{n\in\mathbb{N}}$ converges. Hence the set $\{\phi (s^{k+1}T^{k}) : k\in\mathbb{N}\cup\{0\} \}$ is bounded. As this holds for all $\phi \in \mathcal{B}(X)^{*}$, the set $\{s^{k+1}T^{k} : k\in\mathbb{N}\cup\{0\} \}$ is weakly bounded. It follows from a corollary of the uniform boundedness theorem that the set $\{s^{k+1}T^{k} : k\in\mathbb{N}\cup\{0\} \}$ is norm bounded. Hence there exists some $M \in [0, \infty )$ such that $\|s^{k+1} T^{k}\| \leq M$ for all $k\in\mathbb{N}\cup\{0\}$. Now let $z \in \mathbb{C}$ with $0 < |z| < s$. Then we have $\|z^{k+1} T^{k}\| \leq M (\tfrac{z}{s} )^{k+1}$ for each $k\in\mathbb{N}\cup\{0\}$. As $|\tfrac{z}{s}| < 1$, it follows from completeness that the series $(\sum_{k=0}^{n} z^{k+1} T^{k} )_{n\in\mathbb{N}}$ converges absolutely.