The following might be quite straightforward, but I very rarely work in detail with unbounded operators, so I thought it would be worth seeing quickly if I have overlooked an example that is obvious from the right point of view.


Let $H=L^2(-\infty,\infty)$ and let $S:H \supset {\rm dom}(S) \to H$ be the densely-defined operator $(Sf)(x)=e^xf(x)$, with domain
$$
{\rm dom}(S)=\{ f\in L^2(-\infty,\infty) \colon \int\_{-\infty}^\infty e^{2x} |f(x)|^2\,dx <\infty \}
$$
Then $S$ is self-adjoint.


**Question.** Does there exists a projection $P:H\to H$, with range $V$, satisfying the following properties?


1. $V$ is **infinite-dimensional** and $V':={\rm dom}(S)\cap V$ is dense in $V$.
2. there is a compact operator $R:V\to V$ such that $RPSP\vert\_{V'}= I\_{V'}$?


**Some remarks:**


(a) If we weakened 2. to merely require $R$ being bounded, then this should be easy by taking $V=L^2[0,\infty)$ and $R:V\to V$ to be multiplication by $e^{-x}$. But of course $R$ has continuous spectrum, so it can't be compact.


(b) If we intertwine with the Fourier transform or something similar, perhaps we can we obtain such a $V$ as the solution space to a suitable differential equation? (I have not looked at this vague idea in any detail yet.)


(c) If the answer to the question is positive, can we even get $R$ being trace-class?



Can this self-adjoint operator have an infinite-dimensional compression with compact inverse?
Sure, for instance let $P$ be the orthogonal projection onto the closed span of the characteristic functions $\chi\_{[n,n+1)}$ for $n \in \mathbb{N}$. You get property 1 because each of these functions is in the domain of $S$, and you get property 2 because, identifying $V$ with $l^2$ in the obvious way, the operator $PSP$ becomes multiplication by, I think, $e^{n}(e - 1)$. So this compression has compact, even trace-class inverse, namely multiplication by $e^{-n}(e-1)^{-1}$.


