Can norm of resolvent of closed operator grow away from its spectrum(but near to its numerical range)

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?

• I think I found an answer. Considering the derivative operator $\frac{d}{dx}$ on a bounded interval in a suitable $H^1$ setting with one boundary conditions leads to an empty spectrum but the resolvent grows exponentially in the left half complex plane. Or defining this operator on the whole real line leads to pure continuous spectrum equal to the imaginary axis but the resolvent should grow exponentially in direction of the negative real axis. If there is no answer until I find the time, I will work out one of these examples and post it during the next days. – lsir Apr 1 at 1:19