Resolvent lemma

Question

I would like to proof a lemma that I am quite sure should be correct as I found it somewhere, I am writing a thesis about quantum walks and need this to get through an article.
Let $X$ be Banach space, $A \in B(X)$, $\lambda \in \rho(A)$ (the resolvent set of the operator) I need to proof that:

1. $B(\lambda, \; \frac{1}{||R_{\lambda}||}) \subset \rho(A)$, $B$ is a Ball and $R_{\lambda}$ is the resolvent
2. $dist(\lambda, \sigma(A)) \ge \frac{1}{||R_{\lambda}||}$
   

Thanks, Im am more into the physics and I am not very good at this.

Answer 1

The key ingredient here is the following variant of the von Neumann series.

Proposition: Let $A \in B(X)$ be invertible, and let $B \in B(X)$ be such that $\|BA^{-1}\|<1$. Then $A + B$ is invertible.

Taking this fact for granted, let us see how it implies the desired result. Let $\lambda \in \rho(A)$ and suppose $|\lambda' - \lambda| < \|R_\lambda\|^{-1}$. Then $\|(\lambda' - \lambda)I R_\lambda\| < 1$ so the proposition tell us that $A - \lambda'I = (A - \lambda) + (\lambda - \lambda')I$ is invertible. In other words, $\lambda' \in \rho(A)$.

The second part is a simple consequence of the first because we know that if $\lambda \in \rho(A)$ and $|\lambda - \lambda'| < \|R_\lambda\|^{-1}$ then also $\lambda' \in \rho(A)$. It follows that if $\lambda' \notin \rho(A)$ then $|\lambda - \lambda'| > \|R_\lambda\|^{-1}$. Thus $|\lambda - \lambda'| > \|R_\lambda\|^{-1}$ for all $\lambda' \in \sigma(A)$ which means $d(\lambda, \sigma(A)) > \|R_\lambda\|^{-1}$.

Now let's prove the proposition.

Proof: Define operators $T_N \in B(X)$ by $$T_N =\sum_{n=0}^N (-1)^n A^{-1}(BA^{-1})^n.$$ We will first show that $T_N$ is a Cauchy sequence in $B(X)$. For $N > M$ we have $$\|T_N - T_M\| = \|\sum_{n = M+1}^N(-1)^n A^{-1}(BA^{-1})^n\|$$ $$ \le \sum_{n=M+1}^n \|A^{-1}\| \|BA^{-1}\|^n = \|A^{-1}\|\|BA^{-1}\|^{M+1} \sum_{n=0}^{N-M-1}\|BA^{-1}\|^n$$ Letting $N \to \infty$ and using that the sum is geometric series we get $$\le \|A^{-1}\|\|BA^{-1}\|^{M+1} \frac{1}{1-\|BA^{-1}\|}.$$ Now we let $M \to \infty$ and remember that $\|BA^{-1}\| < 1$ to see that the entire term tends to $0$. Thus $T_N$ is a Cauchy sequence and thus has a limit $T \in B(X)$. We now show that $T$ is a right inverse of $A+B$ which will finish the proof. Notice that $$(A+B)T_N = \sum_{n=0}^N (-1)^n(A+B)A^{-1}(BA^{-1})^n$$ $$= I + \sum_{n=1}^N(-1)^n(BA^{-1})^n + \sum_{n=1}^{N+1} (-1)^n (BA^{-1})^n$$ $$= I + (-1)^N(BA^{-1})^{N+1} \to I$$ as $N \to \infty$.

Since $A+B$ is continuous, this means $$(A+B)Tx = \lim_{N \to \infty} (A+B)T_Nx = x$$ for all $x \in X$ which means $T = (A+B)^{-1}$.