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?


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.)

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  • $\begingroup$ 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 self-adjoint. 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

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