Essential singularity of the resolvent operator of an unbounded operator

Question

Is there an unbounded operator with isolated points in the spectrum, not all of which are eigenvalues? For unbounded operators it is known that isolated spectral points are either poles or essential singularities, and every pole is an eigenvalue. Thus, I need an explicit operator $T$ with $\lambda \in \sigma(T)$ such that $\lambda$ is an essential singularity (and not a pole) of the resolvent operator of $T$. Also, it would be better if someone can give such a point $\lambda$ other than infinity. I'm a little new to the $L^2$ spaces, but I'm quite comfortable working with $\ell^2$ space.

Answer 1

How about a bounded quasinilpotent operator $T \in \mathscr{L}(L^2[0,1])$ defined by $$ (Tf)(x) = \int_{0}^{x}f(t)dt. $$ This operator has $\sigma(T)=\{0\}$. The resolvent cannot have a pole at $0$ because $T$ is not nilpotent of any order. So the resolvent has an essential singularity. You can solve for the resolvent by solving $(T-\lambda I)f = g$ for $f$: $$ (Tf)(x) -\lambda(Tf)'(x) = g \\ (Tf)(0) = 0. $$ To solve this ODE, $$ (Tf)'(x) - \frac{1}{\lambda}(Tf) = -\frac{1}{\lambda}g \\ \frac{d}{dx}\{e^{-x/\lambda}(Tf)(x)\} = -\frac{1}{\lambda}e^{-x/\lambda}g(x) \\ e^{-x/\lambda}(Tf)(x) = -\frac{1}{\lambda}\int_{0}^{x}e^{-t/\lambda}g(t)dt \\ (Tf)(x) = -\frac{1}{\lambda}\int_{0}^{x}e^{(x-t)/\lambda}g(t)dt $$ The resolvent has an essential singularity at $0$, as stated: $$ R(\lambda) g = -\sum_{n=0}^{\infty}\frac{1}{\lambda^{n+1}}\int_{0}^{x}\frac{(x-t)^{n}}{n!}g(t)dt $$ Added for Unbounded Case: Let $\mathcal{H} = L^2[0,1]\times L^2[0,1]$ and define an operator $L(f,g)=(Tf,T^{-1}g)$ on the domain $\mathcal{D}(L)=L^2[0,1]\times\mathcal{R}(T)$. This is an unbounded operator because $T^{-1}$ is unbounded. And $T^{-1}$ has no spectrum. So $L$ is a closed, densely-defined operator with $\sigma(L)=\{0\}$, and the resolvent of $L$ has an essential singularity at $0$.