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Set $ A_i:= -\Delta + V_i :H^2(\mathbb{R}^3) \subseteq L^2(\mathbb{R}^3) \to L^2(\mathbb{R}^3), \ i =1,2 $, where \begin{equation*} V_1 = 0, \ \ (\textrm{No interaction}) \\ V_2 = - \frac{\gamma}{\vert x \vert}, \gamma >0, \ \ (\textrm{Coulumb potential}). \end{equation*} By the fact that $ 0 \in \sigma_c(A_i), \ i =1,2 $,

Remark: This result could be found in Chapter 7, 10 in "G. Teschl: Mathematical Methods in Quantum Mechanics - With Applications to Schrödinger Operators (2014)".

we know \begin{equation*} \mathcal{N}(A_i) = 0, \ \overline{\mathcal{R}(A_i)} = L^2(\mathbb{R}^3) \\ \ A^{-1}_i \ \textrm{unbounded} \ (i =1,2) \end{equation*} Question: Can we find a bounded restriction for $ A^{-1}_i $, that is, there exist $ \mathcal{B}_i \subseteq \mathcal{R}(A_i) $ such that $ A^{-1}_i|_{\mathcal{B}_i} $ are bounded respectively?

Good references will be welcome. Thank you in advance!

Notice: This question is highly related to (Non-closed range space of Laplace operators?) and ($ 0 $ locates in the continuous spectrum of Schrodinger operators?).

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  • $\begingroup$ What kind if conditions do you want $\mathcal{B}_i$ to satisfy? For instance, the situation becomes trivial if you allow finite-dimensional $\mathcal{B}_i$. $\endgroup$
    – Jochen Glueck
    Commented Sep 25, 2019 at 11:19
  • $\begingroup$ @JochenGlueck Now i have no mature thoughts. Maybe you can give me some examples to feel? Even the finite dimensional case will be welcome. Of course the bigger and infinite-dimensional will be better. $\endgroup$
    – Yidong Luo
    Commented Sep 25, 2019 at 11:51
  • $\begingroup$ Well, every linear operator from a finite dimensional space to any normed space is continuous, so you can choose $\mathcal{B}_i$ to be any finite dimensional subspace of $\mathcal{R}(A_i)$ and the restriction of $A_i^{-1}$ to $\mathcal{B}_i$ will be continuous. $\endgroup$
    – Jochen Glueck
    Commented Sep 25, 2019 at 13:09
  • $\begingroup$ @JochenGlueck This is a naive example i ignore. Ok, now the focus is on the infinite-dimensional case. Is there some known research on it? $\endgroup$
    – Yidong Luo
    Commented Sep 25, 2019 at 13:14
  • $\begingroup$ You can, for instance, integrate the spectral measure of your operator over the constant function with value $1$ over any subset of the spectrum that does not contain $0$ in its closure; the result will be a spectral projection $P$ such that the restriction of your operator to the range of $P$ is countinuously invertible. Hence, $\mathcal{B}_i$ can be chosen as the range of any such projection $P$. $\endgroup$
    – Jochen Glueck
    Commented Sep 25, 2019 at 14:24

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