I hope my question is simple to answer but I could not find anything up to now.
Given a closed densely defined operator $A \colon H \supseteq D(A) \to H$ in a Hilbert space $H$. For many types of operators, such as normal ones, one can proof that the resolvent of such an operator $B$ is bounded by $\|(\lambda - B)^{-1}\|_H \leq (\operatorname{dist}(\lambda, \sigma{B})^{-1}$. However, for a general closed densely defined operator I cannot find such a statement but there exist results like $\|(\lambda - A)^{-1}\|_H \leq (\operatorname{dist}(\lambda, W(A))^{-1}$ where $W(A)$ is the numerical range of $A$, i.e. the set $\{ (A u, u)_H \colon u \in D(A), \|u\|_H = 1 \}$.
Now to my question. Given a closed densely defined operator $A$, is it possible that its resolvent grows to infinity, far away from its spectrum $\sigma(A)$?
For example is it possible that $\|(ir - A)^{-1}\|_H \to \infty$ as $ |r| \to \infty$ even though $\sigma(A) \subseteq \mathbb{R}$?
There are operators with real spectrum but with numerical range going far beyond the real axis, so the general result $\|(\lambda - A)^{-1}\|_H \leq (\operatorname{dist}(\lambda, W(A))^{-1}$ suggests to me that the resolvent norm of closed operators is not bounded by the position of their spectrum.
Or is there any kind of theorem/reference for general closed densely defined operators where it is elaborated that far way from the spectrum, the resolvent is bounded?
One example I can give why this is important to me is from the theory of operator semigroups. For example for being an analytic semigroup I need the spectrum in a sector AND a resolvent estimate. If I have the spectrum in the negative real line, I cannot conclude having an analytic semigroup, I also need a resolvent estimate outside a suitable sector right?
Does anyone know a good example, if it exists, for a differential operator where the spectrum is not extending to infinity in the imaginary direction but the resolvent grows to infinity in the imaginary direction?