5
$\begingroup$

Background
This is a follow-up question to: What (classes of) Banach spaces are known to have Schauder basis?

In the previous question, I asked about what spaces are known to have Schauder basis. It seems that not a lot of positive results are available in that area. So I am restricting my question to one of the spaces that is in particular relevant to my research:

Does $C_p$, the Banach space of all Schatten-p operators on a separable infinite-dimensional Hilbert space, have property $\pi$? Or does it have a (Schauder) basis?

Thank you!

Reference
Good reference for different approximation properties: Handbook of the Geometry of Banach Spaces, vol. 1 -- contribution by Pete Casazza

$\endgroup$

1 Answer 1

8
$\begingroup$

It has a Schauder basis, namely $e_i\otimes e_j$, where $(e_i)$ is an orthonormal basis of the Hilbert space. This holds for $1\le p<\infty$. For the Schatten ideal of compact operators, I do not know the answer.

$\endgroup$
3
  • $\begingroup$ I'm not sure if this matters, but it will only be an absolute Schauder basis for $p = 1$, for the other $p > 1$ it is only unconditional, right? $\endgroup$ Jul 11, 2013 at 12:25
  • 5
    $\begingroup$ In natural orders it is a monotone Schauder basis for all $1\le p \le \infty$ (i.e., it is also a Schauder basis for the compact operators); this is elementary. It is unconditional only for $p=2$ (Kwapien-Pelczynski in the 1960s); in fact, for other $p$ there is no unconditional basis at all (Gordon and Lewis in the 1970s). $\endgroup$ Jul 11, 2013 at 15:43
  • $\begingroup$ Thanks for all your help above. It took me a while to figure out what $e_i \otimes e_j$ ("tensorial notation adopted by Schatten"?) means, blame my rusty functional analysis! Definition in p.2 of this article and the fact that all compact operators on Hilbert space are norm limit of finite rank operators are helpful for my understanding: plms.oxfordjournals.org/content/s3-17/1/115.full.pdf and jstor.org/stable/2374892?seq=4 $\endgroup$ Jul 11, 2013 at 21:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.