Let $A, B$ be closable, unbounded, linear operators on a Hilbert space $H$. Suppose further that the operator $A + B$, defined on the intersection of domains $D(A) \cap D(B)$, is also closable.
I would like to show that $\overline{A + B} \supseteq \overline{A} + \overline{B}$.
This is problem 2.8 from Teschl's book Mathematical Methods in Quantum Mechanics.
I have gotten stuck, but I have been trying to rely on the fact that, if $A$ is closable, then the graph $\Gamma(\overline{A})$ is equal to $\overline{\Gamma(A)}$. So if we take $\psi \in D(\overline{A}) \cap D(\overline{B})$. We can get sequences $\{\psi_n\} \subseteq D(A)$, $\{\psi'_n\} \subseteq D(B)$ so that $(\psi_n, A\psi_n) \to (\psi, \overline{A} \psi)$ and $(\psi'_n, B\psi'_n) \to (\psi, \overline{B} \psi)$, where the convergence is in the standard norm on $H^2$, $\|(\psi, \varphi) \|_{H^2} := \|\psi\| + \|\varphi\|$. The problem is that I'd really like to just have one single sequence in $D(A) \cap D(B)$ converging to $\psi$, but I haven't been able to show that this happens.
Hints or solutions are greatly appreciated!
I think that exercise is false.
Let $A=\frac{d^2}{dx^2}$ on all finite linear combinations of $\{ 1,x,x^2,x^3,\cdots \}$ and let $B=\frac{d^2}{dx^2}$ on all finite linear combinations of $\{ \sin(\pi x),\sin(2\pi x),\sin(3\pi x),\cdots\}$. Both of these operators are densely-defined; and they're both closable because they're both restrictions of the same closed operator $C=\frac{d^2}{dx^2}$ on $\mathcal{D}(C)=H^2[0,1]$.
$\mathcal{D}(A+B)=\{0\}$, which means that $A+B$ is trivially closed on its domain $\{0\}$. This is because the linear space of polynomials has nothing in common with finite linear combinations of $\sin(n\pi x)$ terms, other than the $0$ function.
The closure of $B$ has domain $$ \mathcal{D}(\overline{B})=\{ f \in H^2[0,1] : f(0)=f(1)=0 \}. $$ This is because $\{ \sin(n\pi x) \}_{n=1}^{\infty}$ is an orthonormal basis of $\{ f \in H^2[0,1] : f(0)=f(1)=0 \}$.
The domain of $A$ contains polynomials that vanish at $0$ and $1$, which means $\mathcal{D}(A)\cap\mathcal{D}(\overline{B}) \ne \{0\}$. So $\mathcal{D}(\overline{A})\cap\mathcal{D}(\overline{B}) \ne \{0\}$, which proves that $$ \overline{A}+\overline{B} \not \subseteq \overline{A+B}. $$