Given a densely defined symmetric operator $L$ on a Hilbert space $H$, which is also assumed to be diagonalizable, will there always exist a unique extension of $L$ to a selfadjoint operator?

$\begingroup$ Please define precisely what you mean by symmetric diagonalizable operator. $\endgroup$ – Liviu Nicolaescu May 7 '17 at 12:38

$\begingroup$ Symmetric defined as here: en.wikipedia.org/wiki/Extensions_of_symmetric_operators $\endgroup$ – Milan Bernolak May 7 '17 at 12:48

$\begingroup$ . . . . and diagonalizable in the sense that $Dom(L)$ admits a countable basis of eigenvectors of $L$. $\endgroup$ – Milan Bernolak May 7 '17 at 12:50
Yes. If $x_n\in D(L)$ is an ONB of $H$ and $Lx_n=\lambda_n x_n$, then the operator $T$ acting in the obvious way on $D(T)=\{ \sum a_n x_n\in H : \sum \lambda_n^2a_n^2<\infty\}$ is selfadjoint. It is also an extension of $L$ because if $x\in D(L)$, then $$ \lambda_n\langle x, x_n\rangle = \langle Lx, x_n \rangle , $$ so $\sum \lambda_n^2\langle x_n,x\rangle ^2=\Lx\^2<\infty$. In the same way, by approximating $(x,Tx)\in G(T)$ by truncated sums $\sum_{n\le N} a_n x_n\in D(L)$, we see that $T=\overline{L}$, so $L$ is essentially selfadjoint, as required.
(The way I wrote it up this applies to separable Hilbert spaces, but the same argument works in general.)

$\begingroup$ Thanks for the answer! There are a few things not clear to me though. First of all, why is it clear that the domain of the adjoint of $T$ is equal to $D(T)$? $\endgroup$ – Milan Bernolak May 8 '17 at 4:00

$\begingroup$ @MilanBernolak: $T$ is a multiplication operator on $\ell^2$ on its natural domain, these are well known to be selfadjoint. But you can also do it directly, with the same argument that gives you $T\supseteq L$: clearly $T^*\supseteq T$, and if, conversely, $x\in D(T^*)$, then $\lambda_n\langle x,x_n\rangle=\langle T^* x, x_n\rangle$, so $\sum \lambda_n^2\langle x,x_n\rangle ^2 <\infty$ and thus $x\in D(T)$. $\endgroup$ – Christian Remling May 8 '17 at 15:19