Nice Classes of Non-Closable Operators The only thing I know about non-closable operators can be summarised as "they exist, but they're nasty, so let's not talk about them!" This seems to be the case with everyone else I've talked to. I'd appreciate some references or ideas about the following:
What classes of non-closable operators have been studied (if any). For instance have the operators with non-empty, real, positive spectrum been studied? Are there weaker, but similar conditions to closable/symmetric which gaurentee nice properties (e.g. non-empty spectrum, adjoint with non-trivial domain, some form of polar decompostion)?
EDIT: As the comments suggest, the usual definition of spectrum doesn't make sense if the operator's not closable. Is there another way to define it - or is there a similar object which is useful?
 A: Non-closable operators are not necessarily nasty.  Consider the usual Hilbert space $L^2([0,1],dx)$ and the dense subspace $\mathcal{D}=\mathcal{C}[0,1]$.  Define $T$ on $\mathcal{D}$ by $T(f)=f(0)$, i.e. the constant function on $[0,1]$ with value $f(0)$.  This is a densely defined operator, but it is easy to see that its adjoint is not densely defined.  It is not hard to show that in the class of densely defined operators, a closed linear extension exists if and only if the adjoint is densely defined.
A: Here's another simple example of a non-closable operator, from Reed & Simon Vol. I, Section VIII.4, Example 4 (p. 252): Let $D(T)$ consist of those $\psi\in L^2(\mathbf{R})$ such that $\int|f\psi|<\infty$, where $f$ is a fixed function not in $L^2(\mathbf{R})$, and set $T\psi=(f,\psi)\psi_0$ for some fixed $\psi_0 \in L^2(\mathbf{R})$. Then $D(T)$ is dense in $L^2(\mathbf{R})$; however, as they show by a short computation, $D(T^*)$ consists of the vectors orthogonal to $\psi_0$, so it's not dense.
