16
$\begingroup$

Motivation: I am trying to see for what class of Banach spaces the following result is true:

There exists an increasing sequence of finite dimensional subspace {$V_n$} of a Banach space X (with some property) and corresponding projections $P_n: X \to V_n$ such that
a) $\cup V_n$ is dense in $X$
b) $\sup_n ||P_n|| < \infty$

I know if the space has Schauder basis, then the result is automatic. Hence I would appreciate if you can let me know:
i) a reference for the above result (if it exists)
ii) positive results for large class of Banach spaces which has a Schauder basis

I have tried Googling and Wikipedia, but couldn't find systematic information about existence of Schauder basis. The only counterexample I found was given by Per H. Enflo.

Thank you! I apologize if there is any inappropriate etiquette in my post as I am relatively new to the forum.

$\endgroup$
8
  • 3
    $\begingroup$ The property you define is usually called the $\Pi$ property. $\endgroup$ Jun 25, 2013 at 20:12
  • 1
    $\begingroup$ Most of the classical separable Banach spaces are known to have a Schauder basis. The book of Albiac and Kalton is good place to start. Singer's two volumes on bases in Banach spaces probably more than you would ever want to know. $\endgroup$ Jun 25, 2013 at 20:15
  • 2
    $\begingroup$ For a good expository article, read Casazza's contribution in the Handbook of the Geometry of Banach Spaces, vol. 1. $\endgroup$ Jun 25, 2013 at 20:17
  • $\begingroup$ Thanks Bill for pointing me to the references! I will check them out! I have also found this article of you which explains the $\pi$ property: link.springer.com/article/10.1007/BF02771464 $\endgroup$ Jun 26, 2013 at 3:19
  • $\begingroup$ In [JRZ] you will find some results on when the $\pi$ property implies the existence of a finite dimensional decomposition and some results on the existence of bases, such as a separable complemented subspace of an $L+p$ space must have a Schauder basis. $\endgroup$ Jun 26, 2013 at 15:55

1 Answer 1

11
$\begingroup$

The property you define is usually called the $\pi$ property.

For a good expository article, read Casazza's contribution in the Handbook of the Geometry of Banach Spaces, vol. 1.

Most of the classical separable Banach spaces are known to have a Schauder basis. The book of Albiac and Kalton is good place to start. Singer's two volumes "Bases in Banach Spaces" probably contains more than you would ever want to know.

In [JRZ], which you mentioned in a comment, you will find some results on when the $\pi$ property implies the existence of a finite dimensional decomposition and some results on the existence of bases, such as a separable complemented subspace of an $L_p$ space must have a Schauder basis.

$\endgroup$
3
  • $\begingroup$ Thanks for the references given. I have read them except for Singer's two volumes and learnt a lot. I have the impression: the articles available are all trying to show the relation between the different properties -- in particular, whether they are distinct from each other and when a weaker property implies existence of basis. I couldn't find one which illustrate how to show a space has property $\pi$. So, here are my two questions: $\endgroup$ Jul 5, 2013 at 21:50
  • $\begingroup$ 1) Do you know if anyone has considered the relation between Uniform Convex space and $\pi$ spaces? (J Diestel's "Geometry of Banach Spaces - Selected Topics" contains some info about the case when a UC space HAS a basis, while I am trying to see if there are classes of UC spaces known to have a basis) 2) In particular, do you know if $C_p$, the Banach space of Schatten-p class operators on a Hilbert space has property $\pi$? Thank you so much! $\endgroup$ Jul 5, 2013 at 21:54
  • $\begingroup$ Casazza seems to have mentioned a "counterexample" in his handbook article - there exists a reflexive subspace of $l_p$ that fails Compact Approximation Property --> this would be a UC space which fails property $\pi$ $\endgroup$ Jul 5, 2013 at 21:56

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.