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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
$$
rm dom(S)=^2,dx <infty
$$
Then $S$ is self-adjoint.

Question. Does there exists a projection $P:Hto H$, 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 $R:Vto V$ such that $RPSPvert_V'= I_V'$?

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 $R:Vto V$ 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?

Sure, for instance let $P$ be the orthogonal projection onto the closed span of the characteristic functions $chi_[n,n+1)$ for $n in mathbbN$. 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 $e^-n(e-1)^-1$.