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Let $T:\mathrm{dom}(T) \subseteq H \to H$ be a densely defined, self-adjoint operator on a Hilbert spaces $𝐻$. In general the range of $T$ is not guaranteed to be closed. What tools are available to check if the range is closed? More precisely, what are a list of equivalent formulations of closed range, or conditions that imply closed range?

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    $\begingroup$ A hint: thanks to the spectral theorem, you only have to consider multiplication operators on $L^2(X)$... $\endgroup$ – Nate Eldredge Jul 14 '19 at 19:16
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    $\begingroup$ Warning: normally, the term „self-adjoint“ only applies to operators on a single space, i.e. where $H=H‘$. $\endgroup$ – user131781 Jul 14 '19 at 20:04
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    $\begingroup$ I think the current phrasing of the question is too broad in scope, and it would be better to give an example of one of the operators $T$ that you have in mind $\endgroup$ – Yemon Choi Jul 14 '19 at 20:52
  • $\begingroup$ @user131781: I have edited the question. $\endgroup$ – Dave Shulman Jul 14 '19 at 23:10
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This is a complete, but rather abstract characterisation.

$T$ has closed range if and only if there is $H_0\subseteq H$ closed and a bounded self-adjoint injective map $R:H_0\rightarrow H_0$ with $D(T) = \{ \xi+R(\eta) : \xi\in H_0^\perp, \eta\in H_0 \}$ and $T(\xi+R(\eta)) = \eta$.

If this holds, then $T$ is well-defined as $R$ injective, and the image of $T$ is $H_0$ which is closed. The graph of $T$ is $G(T) = \{ (\xi+R\eta, \eta):\xi\in H_0^\perp, \eta\in H_0\}$ which can be shown to be closed. Then $(\alpha,\beta)\in G(T^*)$ if and only if $((-\beta,\alpha)|(\xi+R\eta,\eta))=0$ for $\xi\in H_0^\perp, \eta\in H_0$. This is equivalent to $(\beta|\xi)=(\alpha-R\beta|\eta)$ for all $\eta\in H_0, \xi\in H_0^\perp$, which is equivalent to $\alpha-R\beta\in H_0^\perp, \beta\in H_0^{\perp\perp}=H_0$, that is, $(\alpha,\beta)\in G(T)$. So $T$ is self-adjoint.

Conversely, let $T$ be self-adjoint with closed range $H_0$. Then $H_0 = (\ker T)^\perp$ so $\xi\in H_0^\perp \implies \xi\in D(T), T\xi=0$. Define $S:D(S)\rightarrow H_0$ by $D(S)=D(T)\cap H_0$ and $S(\xi) = T(\xi)$. Then one can show that $S$ is closed, injective, self-adjoint, and onto. Then $R=S^{-1}:H_0\rightarrow H_0$ is closed and everywhere defined (so bounded), self-adjoint, and injective (and so automatically has dense-range). $T$ has the form stated above, using $R$.

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