I am reading Reed-Simon Vol. 4, page 274. We have the following situation: Let $\Omega \subset \mathbb R^m$ be an $m$-dimensional cube and consider the Dirichlet-Laplacian $-\Delta_D=\Delta_D^\Omega$ with quadratic form domain $H^1_0(\Omega)$.
$\textbf{Proposition. }$The set $D_D := \{f: f\in C^\infty(\Omega), f {\restriction}_ {\partial\Omega} = 0 \}$ is an operator core for $-\Delta_D$ and for all $f\in D_D$, $$-\Delta_D f = -\sum_{i=1}^n \frac{\partial^2}{\partial x_i^2} f.$$
In the proof of this proposition they state the following:
Let $A$ be the operator $-\Delta$ with domain $D_D$. $A$ is clearly symmetric, and by separation of variables (multiple Fourier series) we can find a complete orthonormal basis of eigenfunctions $\{ \varphi_n \}$ for A. If $A\varphi_n = \lambda_n \varphi_n$, it is easy to see that $\varphi \in D(\overline{A})$ if and only if $\sum \lambda_n^2 \lvert (\varphi_n, \varphi) \rvert^2 < \infty$; and from this we conclude that $\overline{A}$ is self-adjoint.
Can someone explain this paragraph to me? Two pages later they explicitly give the eigenbasis so I'll count that as an explanation, but the next thing is what I don't see so easily. Why does the "if and only if" hold and why can one conclude that $\overline{A}$ is self-adjoint?
Book excerpt (thanks to Keith McClary).
I thought about this a little bit more and I think I might have an answer. It would be a nice if someone could proofread:
We first note that $D(\overline A) = \overline{D(A)}^{\lVert \cdot \rVert _A}$, where $\lVert \varphi \rVert = \lVert A \varphi \rVert + \lVert \varphi \rVert$. For arbitrary $\varphi \in D(\overline{A}),$ we have by Parseval's equation $$\lVert \overline A \varphi \lVert = \sum_{i=0}^\infty \lvert (\varphi_i, \overline{A} \varphi) \lvert^2 = \sum_{i=0}^\infty \lambda_i^2 \lvert (\varphi_i, \varphi) \lvert^2$$ Now we proof the assertion:
$"\subseteq"$: If for $\varphi \in D(\overline A)$ we have $\sum_{i=0}^\infty \lambda_i^2 \lvert (\varphi, \varphi_i)\rvert^2 = \infty$, by the above $\lVert \overline A \varphi \rVert = \infty$. But this is a contradiction to $\operatorname {im } \overline{A} \subseteq L^2$.
$"\supseteq"$: Assume $\sum_{i=0}^\infty \lambda_i^2 \lvert (\varphi, \varphi_i)\rvert^2 < \infty$ for some $\varphi \in L^2$. We claim that the sequence $$\left( b_n := \sum_{i=0}^n (\varphi_i, \varphi) \varphi_i \right )_{n\in \mathbb N}$$ is a sequence in $D(A)$ and Cauchy with respect to $\lVert \cdot \rVert_A$. Being in $D(A)$ is clear as well as the fact that $\lim_{n\to \infty} b_n = \varphi$ in the $L^2$-norm (since the $\varphi_i$ are an ONB). So it remains to show that $Ab_n$ is a Cauchy sequence. Let $N\in \mathbb N$ and WLOG $m > n \geq N$. We compute $$\lVert A(b_m - b_n) \rVert = \left \lVert A \left(\sum_{k=n+1}^m (\varphi_k, \varphi) \varphi_k\right)\right \rVert \\ = \sum_{l=0}^\infty \lambda_l^2 \left \lvert \left ( \sum_{k=n+1}^m (\varphi_k, \varphi ) \varphi_k ,\varphi_l \right ) \right \rvert^2 \\ = \sum_{l=0}^\infty \left \lvert \sum_{k=n+1}^m \lambda_l (\varphi_k, \varphi) \delta_{kl} \right \rvert^2 = \sum_{l=n+1}^m \lambda_l^2 \lvert (\varphi_l,\varphi) \rvert^2 \leq \sum_{l=N}^\infty \lambda_l^2 \lvert (\varphi_l, \varphi) \rvert^2 \stackrel{N\to \infty}{\longrightarrow} 0.$$ Hence $\varphi \in D(\overline A)$.