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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.

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2 Answers 2

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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.

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  • $\begingroup$ Thanks. I'll have to come back to this when I've studied a little further. $\endgroup$ Apr 11, 2017 at 16:41
  • $\begingroup$ @COVID-20 wonderful answer thank you $\endgroup$
    – kroner
    Oct 9, 2020 at 21:06
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    $\begingroup$ What is a graph inner product? Why is the decomposition orthogonal to the graph inner product? Thank you! $\endgroup$
    – Catcher
    Mar 23, 2021 at 5:12
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    $\begingroup$ @Catcher : The graph norm $\|x\|_{T^*}=\sqrt{\|x\|_H^2+\|T^*x\|_H^2}$ is induced by the graph inner product $\langle x,y\rangle_{T^*}=\langle x,y\rangle_{H}+\langle T^*x,T^*y\rangle_{H}$ $\endgroup$ Mar 23, 2021 at 6:21
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A good background reference would be Rudin's book "Functional Analysis", esp. the information on the Cayley transform and deficiency indices. The following solution is essentially the same as the previous one, just slightly more detailed.

Let $T$ be a densely defined symmetric operator on a Hilbert space $H$. Let $T$ have a unique self-adjoint extension $S$. We wish to show that the closure of $T$ is self-adjoint. Let $C$ be the closure of $T$. We wish to show that $C$ is self-adjoint. Since self-adjoint implies closed, $S$ is an extension of $C$. Since $S$ is the only self-adjoint extension of $T$, it follows that $S$ is the only self-adjoint extension of $C$. (We can now forget about $T$ and focus on $C$ and $S$.)

Let $U:=(C-i)(C+i)^{-1}$ be the Cayley transform of $C$. Let $A$ be the image of $C+i$ and let $B$ be the image of $C-i$. Then (see Rudin) $U:A\to B$ is isometric (i.e., inner product preserving) and bijective. Moreover (see Rudin), because $C$ is closed, it follows that $A$ and $B$ are both closed. Also (see Rudin), since $C$ has a unique self-adjoint extension, $U$ has a unique isometric bijective extension $H\to H$. Finally (see Rudin, vanishing of deficiency indices), to show that $C$ is self-adjoint, it suffices to show that $A=H=B$.

Let $A^\perp$ be the orthogonal complement of $A$ in $H$, and let $B^\perp$ be the orthogonal complement of $B$ in $H$. We wish to show: $A^\perp=\{0_H\}=B^\perp$.

The isometric bijective extensions $H\to H$ of $U$ are in one-to-one correspondence with isometric bijections $A^\perp\to B^\perp$. Then there is a unique isometric bijection $A^\perp\to B^\perp$. In particular, $A^\perp$ and $B^\perp$ are isometric Hilbert spaces.

For any two isometric Hilbert spaces $X$ and $Y$, unless both are zero, there are continuum many isometric bijections $X\to Y$. So, since $A^\perp$ and $B^\perp$ are isometric and only admit one isometric bijection, it follows that both are zero, i.e., that $A^\perp=\{0_H\}=B^\perp$, as desired.

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