I am currently a little bit confused. I am aware of a theorem that says that any closed and densely defined operator satisfies $\sigma(T^*)=\overline{\sigma(T)}.$
On the other hand, the operator
$$T: H_0^1([0,1]) \subset L^2([0,1]) \rightarrow L^2([0,1]), f \mapsto if'$$
has the adjoint operator
$$T^*: H^1([0,1]) \subset L^2([0,1]) \rightarrow L^2([0,1]), f \mapsto if'.$$
The first one (T) has empty spectrum while its adjoint $T^*$ has $\sigma(T^*)=\mathbb{C}.$ Despite, both operator are closed and densely defined, so somehow this result is not reconcilable with the theorem I just stated. Is the theorem wrong or my example?
If anything is unclear, please let me know.
The first operator $T$ does not have empty spectrum. To see why, suppose you want to solve $(T-\lambda I)f=g$ for $f \in H_{0}^{1}$. Then $$ if'-\lambda f=g \\ f'+i\lambda f =-i g\\ (e^{i\lambda t}f)' = -ie^{i\lambda t}g \\ \int_{0}^{1}(e^{i\lambda t}f)'dt = -i\int_{0}^{1}e^{i\lambda t}g(t)dt \\ 0=e^{i\lambda t}f|_{0}^{1} = -i\int_{0}^{1}e^{i\lambda t}g(t)dt $$ Therefore, there is no chance of solving the above unless $(g,e^{-i\lambda t})=0$. That means the range of $T-\lambda I$ is always deficient, for all $\lambda\in\mathbb{C}$, which means $\sigma(T)=\mathbb{C}$.