Let $T:\mathrm{dom}(T) \subseteq H \to H$ be a densely defined, selfadjoint 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?

1$\begingroup$ 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

4$\begingroup$ Warning: normally, the term „selfadjoint“ only applies to operators on a single space, i.e. where $H=H‘$. $\endgroup$ – user131781 Jul 14 '19 at 20:04

1$\begingroup$ 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
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 selfadjoint 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 welldefined 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 $(\beta\xi)=(\alphaR\beta\eta)$ for all $\eta\in H_0, \xi\in H_0^\perp$, which is equivalent to $\alphaR\beta\in H_0^\perp, \beta\in H_0^{\perp\perp}=H_0$, that is, $(\alpha,\beta)\in G(T)$. So $T$ is selfadjoint.
Conversely, let $T$ be selfadjoint 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, selfadjoint, and onto. Then $R=S^{1}:H_0\rightarrow H_0$ is closed and everywhere defined (so bounded), selfadjoint, and injective (and so automatically has denserange). $T$ has the form stated above, using $R$.