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 self-adjoint operator?



Symmetric diagonalizable operators and self-adjointness
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 $D(T)=\{ \sum a\_n x\_n\in H : \sum \lambda\_n^2|a\_n|^2<\infty\}$ is self-adjoint. 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 self-adjoint, as required.


(The way I wrote it up this applies to separable Hilbert spaces, but the same argument works in general.)


