Prove the following inequality on a Banach Algebra $\mathcal{A}$ Let $x \in \mathcal{A}$ and $K \subset \mathbb{C}$. Let $r$ be the minimum distance of $K$ from $σ(x)$ (where $σ(x)$ is the spectrum of $x$) and also $r > \delta$ for some $\delta > 0$. Show that there exists a $M>0$ such that $||(λe-x)^{-n}|| \leq M \delta^{-n}$, for all $n \in \mathbb{N}$ and $λ \in K$.
Thanks for any help.
 A: Since
$$
\lim_{n\to\infty}\|(\lambda e-x)^{-n}\|^{1/n}=\operatorname{spr}\big((\lambda e-x)^{-1}\big),
$$
there exists $n_0$ such that
$$
\|(\lambda e-x)^{-n}\|^{1/n}\leq\tfrac{\delta^{-1}-r^{-1}}2+\operatorname{spr}\big((\lambda e-x)^{-1}\big),\qquad n\geq n_0. 
$$
Then, for $n\geq n_0$,
\begin{align}
\|(\lambda e-x)^{-n}\|^{1/n}&\leq
\tfrac{\delta^{-1}-r^{-1}}2+\operatorname{spr}\big((\lambda e-x)^{-1}\big)
=\tfrac{\delta^{-1}-r^{-1}}2+\sup\{|\alpha|:\ \alpha\in\sigma\big((\lambda e-x)^{-1}\big)\}
\\[0.3cm]
&=\tfrac{\delta^{-1}-r^{-1}}2+\sup\Big\{\tfrac1{|\alpha|}:\ \alpha\in\sigma(\lambda e-x)\Big\}\\[0.3cm]
&=\tfrac{\delta^{-1}-r^{-1}}2+\frac1{\inf\{|\alpha|:\ \alpha\in\sigma(\lambda e-x)\}}\\[0.3cm]
&=\tfrac{\delta^{-1}-r^{-1}}2+\frac1{\inf\{|\lambda-\alpha|:\ \alpha\in\sigma(x)\}}\\[0.3cm]
&\leq\tfrac{\delta^{-1}-r^{-1}}2+\frac1r<\delta^{-1}.
\end{align}
Let
$$
M=1+\max\{\delta^n\|(\lambda e-x)^{-n}\|:\ n=1,\ldots,n_0\}.
$$
Then
$$
\|(\lambda e-x)^{-n}\|\leq M\delta^{-n}
$$
for all $n$ and any $\lambda\in K$.
