Is it possible that $\|(z-a)^{-1}\|\gneq r((z-a)^{-1})$ for some $z\notin \sigma(a)$ but $\|(z-a)^{-2}\|= r((z-a)^{-2})\ \forall z\notin \sigma(a)$?

Question

Let $A$ be a Banach algebra and $a\in A$. In general, $\|(z-a)^{-1}\|\geq \|(z-a)^{-2}\|^{\frac{1}{2}}\ \forall z\notin \sigma(a)$.

Is it possible that

$\|(z-a)^{-1}\|\gneq r((z-a)^{-1})$ for some $z\notin \sigma(a)$ but

$\|(z-a)^{-2}\|= r((z-a)^{-2})\ \forall z\notin \sigma(a)$?

Is there a concrete example of this? For instance, for any non-normal matrix with the induced Euclidean norm,

$\|(z-a)^{-1}\|\gneq r((z-a)^{-1})$ for some $z\notin \sigma(a)$. Is it possible to have a non-normal matrix for which

$\|(z-a)^{-2}\|= r((z-a)^{-2})\ \forall z\notin \sigma(a)$?

I'd be grateful for a hint on how to begin.

Here, $r(b)$ denotes the spectral radius of an element $b\in A$.

Answer 1

Let $A=M_2(\mathbb C)$, and $$ a=\begin{bmatrix}2&1\\0&0\end{bmatrix}. $$ Then $\sigma(a)=\{0,2\}$. Take $z=1$. Then $$ z-a=\begin{bmatrix}-1&-1\\ 0&1\end{bmatrix},\ \ \ (z-a)^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}. $$ Thus $$ (z-a)^{-1}=\begin{bmatrix}-1&-1\\ 0&1\end{bmatrix},\ \ \ (z-a)^{-2}=\begin{bmatrix}1&0\\0&1\end{bmatrix}, $$ and $$ \|(z-a)^{-1}\|=\sqrt{\frac{3+\sqrt5}2},\ \ \ r((z-a)^{-1})=1, $$

$$ \|(a-z)^{-2}\|=1,\ \ \ r((z-a)^{-2})=1. $$