Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

... are all isomorphic to $l^1$ on some other index set. At least, that much I "know" from 2nd-hand sources, since the original proof is apparently in a paper of Köthe from the 1930s 1960s (in German) that I can't get hold of have had trouble digesting. Since there are some Banach space specialists reading MO, I wondered if someone could sketch how the proof differs from the countable case, or point to a more recent text, preferably in English or French, that gives the proof? I hope this is a well-defined question for MO, since I'm not baiting with something where I know the answer.

(Some background for other readers: the analogous result when $X$ is countably infinite follows from combining two steps: one first uses a block basis argument to show that a closed, complemented subspace $V$ inside $l^1(\bf N)$ must either be finite-dimensional, or contain an infinite-dimensional, closed complemented subspace $W$ that is isomorphic to $\ell^1({\bf N})$. In the former case, $V$ is then obviously isomorphic to some finite-dimensional $\ell^1$. In the latter case, one applies Pelczynski decomposition. My impression is that it's the first step which might prove problematic if attempted for $\ell^1(X)$ when $X$ is uncountably infinite, but I could well be wrong and would welcome corrections.)

share|improve this question
3  
I don't read German, but according to arxiv.org/abs/math/0501048 Köthe did this in springerlink.com/content/u58k505417h14214, published in 1966 and hopefully accessible to you as it is to me via institutional login. I don't know an English or French source. –  Jonas Meyer Nov 15 '09 at 2:22
    
Ah, I misremembered the date, but that's the title of the paper I was thinking of. Hmm, didn't realise it was now on SpringerLink (it's been a while since the original "need" I had for it arose). Thanks, although if anyone reading can give me an explanation of the relevant part that'd be great. –  Yemon Choi Nov 15 '09 at 2:59
add comment

2 Answers 2

up vote 7 down vote accepted

A proof in English (modulo some details involving the Pelczynski decomposition method) can be found in the article 'On relatively disjoint families of measures' (Studia Math, 37, p.28-29) by Haskell Rosenthal.

Regarding the analogous result for $\ell_p (X)$ ($p\in (1,\infty)$) and $c_0 (X)$, I seem to recall reading somewhere that it was solved by Joram Lindenstrauss, but now I can't seem to find any reference to it. I seem to think that I saw something about it in the Appendix to the English translation of Banach's book on linear operations (the appendix is by Bessaga and Pelczynski), but it would take me a while to sift through it to find it, and family dinner is being dished up very shortly. I wonder anyhow how much can be gleaned from Matthew Daws' classification of the closed, two-sided ideals in $\mathcal{B}(\ell_p (X))$, the Banach algebra of all bounded linear operators on $\ell_p (X)$? The relevant paper can be downloaded at http://www.amsta.leeds.ac.uk/~mdaws/pubs/ideals.pdf . The paper 'The lattice of closed ideals in the Banach algebra of operators on a certain dual Banach space' by Laustsen, Schlumprecht and Zsak illustrates how classification of the complemented subspaces of a Banach space $E$ can follow from the classification of closed, two-sided ideals in $\mathcal{B}(E)$ if all the closed, two-sided ideals are generated by projections onto complemented subspaces having certain nice properties. How much of this can be done using Matt's results I haven't checked, but I think that at the very least some partial results could be obtained. Matt might comment of this if he passes by, or if no one else does I might try to look into it in the next day or so and edit this answer accordingly.

The analogous result for $\ell_\infty (X)$ does not hold for uncountable $X$ in general. Indeed, every $\ell_\infty (X)$ is the dual of $\ell_1 (X)$, every Banach space embeds isomorphically into some $\ell_\infty (X)$, but there are injective Banach spaces that are not isomorphic to any dual Banach space; the first such example seems to have been found by Haskell Rosenthal in his paper 'On injective Banach spaces and the spaces $L^\infty (\mu)$ for finite measure $\mu$' (Acta Mathematica, 124, Corollary 4.4), and the existence of such a space provides the desired counterexample.

share|improve this answer
1  
Excellent! Thanks Phil - I didn't think to look in the Rosenthal paper, but on a quick look it seems just what I wanted. (The original motivation, back in 2003-04, was that Kothe's result tells us there are no interesting projective Banach spaces, since they'd all have to be complemented subspaces of some l^1(X). At one point I might have wanted/needed to use the result and was reluctant to do so, since I'd never read the proof.) –  Yemon Choi Nov 15 '09 at 8:47
add comment

Nice answer, Phil.

The case of $Z=\ell_p(X)$, 1 < p < infinity, and $Z=c_0(X)$ with $X$ uncountable are a bit easier because if $T$ is a bounded linear operator from $Y$ into $Z$ and the density character of $Y$ is smaller than $|X|$, then $T$ cannot be one to one. So if $Y$ is a subspace of $Z$ with density character $|X|$, then a maximal set of disjointly supported unit vectors in $Y$ has cardinality $|X|$ (use the obvious fact that a subspace $W$ of $Z$ is contained in $\ell_p(Q)$ [or $c_0(Q)$] with $|Q|$ equal to the density character of $W$). So every subspace of $Z$ with density character $|X|$ has a norm one complemented subspace that is isometric to $Z$. This and a second use of the ``obvious fact" gives the desired result.

share|improve this answer
    
Thanks for sketching the details of the proof Professor Johnson. I figured it would be something like that, but I didn't have time to nut it out when I wrote my answer above. Oh, and thanks for the compliment on my answer! :) –  Philip Brooker Apr 9 '10 at 11:20
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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