Is it possible to classify non-closed subspaces of Hilbert's space?

Question

Let $H$ be Hilbert's space.

Motivated by my previous question about wildly discontinuous linear functionals, which may be interpreted as an attempt to classify dense hyperplanes in $H$, let me now go straight to the point:

Questions.

1. Are there any significant differences among dense hyperplanes in $H$?

2. If $L$ and $M$ are two dense hyperplanes in $H$, is there a unitary operator mapping $L$ to $M$?

3. Assuming the answer to (2) is negative, how many orbits are there for the natural action of the unitary group $\mathscr U(H)$ on the set of dense hyperplanes?

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Speaking about general (not necessarily closed or dense) subspaces of $H$, there are a few things one may say in that respect.

For example, not all such spaces may be described as the range of a bounded operator and, in particular, no dense hyperplane qualifies. This is because, if the range of such an operator has finite co-dimension, it must be closed (this follows easily from the Closed Graph Theorem).

The range of a compact operator does not contain any infinite dimensional closed subspace, so that is another property one could use to classify subspaces.

More Questions.

4. Is there a necessary and sufficient condition, expressed in topological/analytical terms, characterizing the range of a bounded (resp. compact) operator among all subspaces of $H$?

5. How many unitary equivalence classes of non-closed subspaces of $H$ are there? How many of these may be described in topological/analytical terms?

Answer 1

I guess I have a simple answer to Question 4, in the compact case: An infinite dimensional subspace $E\subseteq H$ is the range of a compact operator iff there exists an orthogonal (as opposed to orthonormal) set $\{e_n\}_{n\in {\mathbb N}}\subseteq E$, such that $$ \lim_{n\to \infty }\Vert e_n\Vert = \infty , $$ and $$ E=\Big\{\xi \in \overline{\text{span}\{e_n\}}: \sum_{n=1}^\infty \big|\langle \xi , e_n\rangle \Big|^2<\infty \Big\}. $$ This follows easily from the Spectral Theorem for compact operators, and the fact that the range of a compact operator $T$ coincides with the range of $|T|$.