In my Functional Analysis course, we proved that for a (possibly unbounded) operator $T$ that is densely defined, closed, and symmetric, exactly one of the following four occurs:
- $\sigma(T) = \mathbb C$;
- $\sigma(T) = \{\lambda \in \mathbb C \mid \Im \lambda \geq 0\}$;
- $\sigma(T) = \{\lambda \in \mathbb C \mid \Im \lambda \leq 0\}$;
- $\sigma(T) \subset \mathbb R$.
Now, 4 is easy; this is true for selfadjoint operators. I'm having a hard time coming up with an example for option 1; can you guys help me?
Let $T=\frac{1}{i}\frac{d}{dt}$ be defined on the domain $\mathcal{D}(T)$ consisting of all absolutely continuous functions $f \in L^2[0,1]$ for which $f(0)=0=f(1)$. More precisely, $f \in \mathcal{D}(T)\subset L^2[0,1]$ is an equivalence class of functions equal a.e. with one element $\tilde{f}$ of the equivalence class that is absolutely continuous on $[0,1]$ with $\tilde{f}'\in L^2[0,1]$. Then $T$ is closed and densely-defined. It's not hard to check that $T$ is symmetric: $$ (Tf,g)-(f,Tg) = \frac{1}{i}\int_{0}^{1}f'\overline{g}+f\overline{g}'dt=\left.\frac{1}{i}f\overline{g}\right|_{0}^{1} = 0. $$ The resolvent equation is $(T-\lambda I)f=g$, which means $$ f'-i\lambda f=ig,\;\;\; f(0)=0=f(1). $$ Using an integrating factor $e^{-i\lambda t}$ and the fact that $f(0)=0$ must hold, you can see that the following is necessary: $$ \frac{d}{dt}(fe^{-i\lambda t})=e^{-i\lambda t}ig \\ f(x)e^{-i\lambda x} = \int_{0}^{x}e^{-i\lambda t}ig(t)dt $$ However, this is an actual solution iff $$ \int_{0}^{1}e^{-i\lambda t}g(t)dt = 0. $$ So there is no $\lambda\in\mathbb{C}$ for which a solution of the resolvent equations can be found for all $g$. Therefore $\sigma(T)=\mathbb{C}$.