finite rank approximation on separable Hilbert spaces
Theorem Let be a separable Hilbert space and let . Then is a compact operator![]()
iff there is a sequence of finite rank operators with .
Proof.
: Assume is compact on and is an orthonormal basis of . Define:
It is clear that the have finite rank and that we have for all , .
Let be the unit ball in . We have that pointwise. Since the are contractive they are equicontinuous, hence converges uniformly to on compact sets, and in particular on , which is compact by assumption.
Therefore uniformly on , hence .
Since is bounded
and of finite rank the first direction follows.
: Now let be a sequence of bounded operators![]()
of finite rank with .
We have to show that is relatively compact in . This is equivalent
![]()
to being totally bounded
in .
So we are left to show that for all there is an -net so that:
So choose and fixed so that:
Choose with:
Hence (by the triangle inequality):
and we are done. ∎
| Title | finite rank approximation on separable Hilbert spaces |
|---|---|
| Canonical name | FiniteRankApproximationOnSeparableHilbertSpaces |
| Date of creation | 2013-03-22 18:23:18 |
| Last modified on | 2013-03-22 18:23:18 |
| Owner | karstenb (16623) |
| Last modified by | karstenb (16623) |
| Numerical id | 10 |
| Author | karstenb (16623) |
| Entry type | Theorem |
| Classification | msc 46B99 |