Is there any relation between $||(z-A)^{-1}||$ and $||(\frac{1}{z}-A^{-1})^{-1}||$ for $z \in \rho(A) \setminus \{0\}$?

Question

Let $A: D(A) \to X$ be an operator defined in a Banach space $X$ which has bounded inverse. I know that if $z \in \rho(A) \setminus \{0\}$, then $\frac{1}{z} \in \rho(A^{-1})$. But, is there any relation between $||(z-A)^{-1}||$ and $||(\frac{1}{z}-A^{-1})^{-1}||$ for $z \in \rho(A) \setminus \{0\}$?

Answer 1

Since $\frac{1}{z}-A^{-1}=\frac{1}{z}A^{-1}(A-z)$, $$\|(\frac{1}{z}-A^{-1})^{-1}\|=\|z(z-A)^{-1}A\|\le|z|\|A\|\|(z-A)^{-1}\|$$