Let $H$ be a Hilbert space and $A \in \mathcal{C}(H)$ a densely defined, closed operator (so possibly unbounded) and $A^*$ its adjoint operator. Now given the condition that $A^2 = 0$, i.e. $\mathop{\mathrm{im}} A \subseteq \ker A$, the Hodge operator $$D := A + A^*$$ defined on $\mathop{\mathrm{dom}} D = \mathop{\mathrm{dom}} A ∩ \mathop{\mathrm{dom}} A^*$ is again densely defined, closed and also self-adjoint.
The question is now, whether this lemma still holds without requiring that $A^2 = 0$. That $D$ is symmetric is obvious, but I couldn't show that $\mathop{\mathrm{dom}} D = \mathop{\mathrm{dom}} D^*$. Is there maybe a good counter example?
A silly counterexample: Let $A$ be closed, symmetric but not self-adjoint. Then $A \subsetneqq A^\ast$, so $D = 2A$ is closed, symmetric but not self-adjoint. Take e.g. $A = i\frac{d}{dt}$ on the set of absolutely continuous functions with zero boundary values and square-integrable derivative on $L^2[0,1]$. – t.b. Jun 18 '12 at 13:06
It's even worse: If $A^2\neq 0$ then in general $D$ is not even densely defined so it doesn't even have a (unique) adjoint. For example there are closed operators $A$ with $\mathrm{dom}A\cap\mathrm{dom}A^*=\{0\}$... – Freeze_S Jun 12 '14 at 12:39