I have two definitions of an essentially self adjoint operator

  1. A symmetric operator with a self adjoint closure
  2. An operator with a unique self adjoint extension.

I can easily show that (1) implies (2) but the converse escapes me. I can get as far (correctly I hope) that if $T$ has a unique self adjoint extension $S$ then $T$ is densely defined, closable, and symmetric and $ T \subset cl(T) \subset S = S^* \subset [cl(T)]^* = T^*$

I have a reference Reed & Simon, V.1 p.256 which further refers to section (V.2) X.1, but I didn't find the answer there.


Suppose $T$ is symmetric, and has a unique selfadjoint extension. (I'll assume that $T$ is densely-defined, though you can probably prove that it must be the case.)

Let $\overline{T}$ be the closure of $T$, which exists because $T$ is symmetric. $\overline{T}$ must also be symmetric, and because $\overline{T}^*=T^*$, then $$ \mathcal{D}(T^*)=\mathcal{D}(\overline{T})\oplus\mathcal{N}(T^*-iI)\oplus\mathcal{N}(T^*+iI), $$ where the decomposition is orthogonal with respect to the graph inner product $$ (x,y)_{T^*} = (x,y)_H+(T^*x,T^*y)_{H}. $$ If $S$ is a selfadjoint extension of $T$, then $S$ is also a selfadjoint extension of $\overline{T}$ because the graph of $T$ is contained in the graph of $S$ and the graph of $S$ is closed. $\overline{T}$ has a selfadjoint extension iff $\mathcal{N}(T^*-iI)$ is unitarily equivalent to $\mathcal{N}(T^*+iI)$, and the possible selfadjoint extensions are in one-to-one correspondence with the unitary maps $U: \mathcal{N}(T^*-iI)\rightarrow\mathcal{N}(T^*+iI)$. However, because $\overline{T}$ has only one selfadjoint extension, then $\mathcal{N}(T^*-iI)=\{0\}=\mathcal{N}(T^*+iI)$ follows. Hence, $\overline{T}=T^*$, which also gives $\overline{T}=\overline{T}^*$. So $\overline{T}$ must be selfadjoint.

  • $\begingroup$ Thanks. I'll have to come back to this when I've studied a little further. $\endgroup$ – Tom Collinge Apr 11 '17 at 16:41

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