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Let $T$ be a symmetric (not necessarily densely defined) (unbounded) operator in a Hilbert space $H$. I want to show, using the theory of the Cayley transform (as developed in e.g. grandpa Rudin), that $T$ has a symmetric closed extension. That this holds is claimed on p339 of the aforementioned book.

I tried as follows:

Let $U$ be the Cayley transform of $T$. Then $U: D(U) \to H$ is isometric, and it therefore extends uniquely to an isometry $$\overline{U}: \overline{D(U)}^{\|\cdot\|} \to H.$$

What I now aim to prove is that $I-\overline{U}$ is injective, because in that case (by theorem 13.9 in grandpa Rudin), there will exist a symmetric operator $\overline{T}$ such that $\overline{U}$ is the Cayley transform of $\overline{T}$. Since $\overline{U}$ is a closed operator, the same is true for $\overline{T}$ and then $\overline{T}$ is the desired extension of $T$. However, I am not sure why $I-\overline{U}$ is injective. Note that we know that $I-U$ is injective.

Any help will be appreciated!

Andromeda
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  • What is grandpa Rudin? ^^ – Severin Schraven Jan 15 '22 at 19:02
  • Rudin's third book (Functional analysis). See https://math.stackexchange.com/questions/1863512/baby-papa-mama-big-rudin – Andromeda Jan 15 '22 at 19:17
  • That is so funny! I heard baby Rudin before, but not the others. Thanks for sharing! – Severin Schraven Jan 15 '22 at 19:34
  • @Severin Schraven You are welcome! – Andromeda Jan 15 '22 at 19:38
  • What is the argument that $I-U$ is injective? Maybe we can adapt that proof – Severin Schraven Jan 15 '22 at 19:44
  • @SeverinSchraven If $(I-U)z=0$ for $z \in D(U)= \operatorname{Im}(T+i I)$, write $z= Tx+ix$ for $x \in D(T)$ and note that $0 = (I-U)z = (Tx + ix)-(Tx-ix) = 2ix$ so $x=0$ and thus also $z=0$. Recall that $U$ is defined to be the isometry $D(U):=\operatorname{Im}(T+iI) \to \operatorname{Im}(T-iI): Tx + ix \mapsto Tx-ix$ where the isometric property holds because $T$ is symmetric. – Andromeda Jan 15 '22 at 19:49
  • @Severin Schraven Based on your suggestion, I quickly tried the following: Suppose $(I-\overline{U})z = 0$ for $z \in \overline{D(U)}$. Write $z= \lim_n z_n$ where $z_n \in D(U)$. Then because $I-\overline{U}$ is continuous on $\overline{D(U)}$ we get $$0 = \lim_n (I-U)z_n = \lim_n 2i x_n$$ where $x_n$ is chosen such that $z_n = Tx_n + ix_n$. We thus have $x_n \to 0$, but unfortunately we cannot guarantee that $z_n \to 0$ because we don't know if $Tx_n \to 0$ ($T$ is unbounded). – Andromeda Jan 15 '22 at 19:56
  • Hmmm, I see. It is weird that it should be so hard to show that symmetric operators are closable, given that it is so easy in the densily defined case. – Severin Schraven Jan 15 '22 at 20:54
  • @SeverinSchraven Yeah, Rudin spends only 2 sentences about it, so a bit odd that it doesn't work out after some effort. Perhaps I'm missing something obvious though. – Andromeda Jan 15 '22 at 21:10
  • I mean the standard notion of symmetric operators assumes that it is densely defined. Maybe he implicitely assumed that. In that case it is enough to know that $x_n\rightarrow 0$ and $a=\lim_{n\rightarrow \infty} Tx_n$ exists to conclude that $a=0$. For this consider any $z\in D(T)$, then we have $$\langle a, z\rangle =\lim_{n\rightarrow \infty} \langle x_n, Tz\rangle =0.$$ As this holds for a dense set of $z$, we get by the Riesz representation theorem that $a=0$. – Severin Schraven Jan 15 '22 at 22:01
  • @SeverinSchraven Thanks yes I guess that makes sense. You don't need Riesz representation theorem though, just continuity of the inner product. – Andromeda Jan 15 '22 at 22:07
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    Ok, yes, for me the step ($\langle a, z\rangle =0$ for a dense set of $z$ implies $a=0$) is Riesz, but of course we could use continuity to conclude $\langle a, z\rangle =0$ for all $z$ and so also $\Vert a\Vert^2=\langle a, a\rangle =0.$ – Severin Schraven Jan 15 '22 at 22:15

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