# Characterising closed range self-adjoint operators

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?

• A hint: thanks to the spectral theorem, you only have to consider multiplication operators on $L^2(X)$... – Nate Eldredge Jul 14 '19 at 19:16
• Warning: normally, the term „self-adjoint“ only applies to operators on a single space, i.e. where $H=H‘$. – user131781 Jul 14 '19 at 20:04
• 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 – Yemon Choi Jul 14 '19 at 20:52
• @user131781: I have edited the question. – Dave Shulman Jul 14 '19 at 23:10

$$T$$ 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 $$R$$.