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?

  • 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 „self-adjoint“ 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 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 $R$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.