I noticed something when working on a problem in Hall's Quantum Theory for Mathematicians. The original problem is that if $A \equiv -i\hslash \frac{d}{dx}$ on $\mathcal{H} = L^2[0,1]$ and
$$ Dom(A) \;\; =\;\; \{\psi\in\mathcal{H} \; | \; \psi \; \text{smooth}, \psi(0) = \psi(1) = 0\} $$
Noting that the closure and the adjoint are distinct operators with $Dom(A^{cl})$ being those continuous $\psi \in \mathcal{H}$ with the same boundary conditions in $Dom(A)$, and $Dom(A^*)$ includes all smooth elements of $\mathcal{H}$.
The original problem was to show that $\sigma(A^{cl}) = \mathbb{C}$ having already proven that $\sigma(A^*) = \mathbb{C}$.
My argument is fairly straightforward since each domain $Dom(A), Dom(A^{cl}), Dom(A^*)$ is dense in $\mathcal{H}$ by definition. Thus for any $\psi \in Dom(A^*)$ of the form $\psi(x) = e^{i\lambda x/\hslash}$, we can find a sequence of unit vectors $\psi_n \in Dom(A^{cl})$ such that $\psi_n \to \psi$ and $||(A^{cl} - \lambda I)\psi_n|| \to 0$, proving that $\lambda \in \sigma(A^{cl})$ and $\lambda$ chosen arbitrarily from $\mathbb{C}$.
I'm a little surprised by this because I get the sense the same construction can be used to prove that $\sigma(A^{cl}) = \sigma(A^*)$ for any symmetric operator $A$. Is this true, or is this statement an over-generalization?