Let $X$ be a Banach space and $T:D(T)\subseteq X \to X$ be a closed operator such that $\sigma(T)= \emptyset$. Let $\epsilon>0$. I need to show that there exists $z_0 \in \mathbb{C}$ such that $\|(z_0-T)^{-1} \|> \epsilon$.
My attempt:
I suppose that $ \|(z-T)^{-1} \| \leq \epsilon$ for all $ z \in \mathbb{C}$. So the function $z \to (z-T)^{-1} $ is a bounded entire function and by Liouville theorem there exists a bounded operator $A:X \to X$ such that $(z-T)^{-1}=A$ for $ z \in \mathbb{C}$. In particular, $ 1-T=A^{-1}=0-T$ (contradiction).
Is my attempt correct?