# Can this self-adjoint operator have an infinite-dimensional compression with compact inverse?

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?

• "Projection" here means "orthogonal projection"? – Nate Eldredge Aug 17 '18 at 3:09
• @NateEldredge oops, yes. (Had my op alg hat on, where that is the usual shorthand) – Yemon Choi Aug 17 '18 at 3:25

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 $e^{-n}(e-1)^{-1}$.
• BTW, by comprssing to an increasing sequence $(n_k)$ you can make the eigenvalues of $PSP$ go to infinity as fast as you want. – Nik Weaver Aug 17 '18 at 13:52