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I was trying to find papers and articles about non-contractive continuous projections on $\ell_1(S)$ where $S$ is an arbitrary set. If it is not studied yet, I would like to know results for the case $S=\mathbb{N}$.

I've found one quite general condition for the closed linear subspace to be image of a continuous projection. Such a subspace must be the closure of the linear span of so-called relatively disjoint vectors. This subspaces gives us explicit examples of projections with norm greater than 1. For details see the paper of H. P. Rosenthal On relatively disjoint families of measures, with some applications to Banach space theory.

As a special case we can get subspaces that are the closure of the linear span of disjointly supported vectors. These subspaces give us examples of norm one projections. Moreover, only such subspaces are give rise to norm one projections. For details see the survey by Beata Randrianantoanina Norm one projections in Banach spaces.

Thus, for projections of norm 1 we have a complete description. For the rest quite a big source of examples. In the first mentioned paper the author states that he doesn't know any other examples of continuous projections on $\ell_1(S)$ that are not generated by some relatively disjoint family of vectors.

So, could someone give me a reference where I can read about other examples of projections on $\ell_1(S)$, or may be their complete characterization?

Also I will be grateful if you give me some explicit examples of discontinuous projections on $\ell_1(S)$.

The same question on mathoverflow.net.

freishahiri
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Norbert
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    If you put a 500 bounty on this its chances of getting answered will sky rocket. – Ragib Zaman Jun 10 '12 at 15:00
  • Are you interested in (1) subspaces which don't admit continuous projections [namely, uncomplemented subspaces], or (2) in discontinuous projections themselves? The examples of (2) can be constructed even in a Hilbert space, where all subspaces are complemented. –  Jun 16 '12 at 18:56
  • I'm interested in quite explicit description of non-contractive projections. I've will have it, of course, I'll get examples of discontinuous projections. Examples of uncomplemented subspaces will be helpful too. – Norbert Jun 16 '12 at 19:30
  • Thanks @JonasMeyer, I wanted to put this bounty in the end of the summer, but you are the first. If someone answer it (the way I wanted) I'll put another 500 bounty for him/her. – Norbert Jun 27 '12 at 22:31
  • Here's an idea (which I leave as a comment rather than an answer since I have not checked enough of the details). Take a subspace $E$ of $\ell^1$ that is isomorphic with small Banach-Mazur constant to $\ell^2$. (You can probably get something out of Khintchine's inequality.) Take a large finite-dimensional subspace of $E$, call it $F$. Then any projection of $\ell^1$ onto $F^\perp$ should have large norm, since otherwise by taking $F$ larger and larger we could eventually build a projection of $\ell^1$ onto $E^\perp$ and hence onto $E$, which is impossible. –  Jul 01 '12 at 09:41
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    @Yemon: I don't understand your comment. There is no subspace of $\ell^1$ isomorphic to $\ell^2$ by Pitt's theorem. A closed infinite-dimensional subspace of $\ell^p$ cannot be isomorphic to a closed subspace of $\ell^q$ if $1 \leq p \neq q \lt \infty$ since the operator in one direction would have to be compact. – t.b. Jul 02 '12 at 12:14
  • @Norbert: Do you search for a construction of a discontinuous projection in $\ell_1(S)$, without using the axiom of choice ? – francis-jamet Jul 02 '12 at 13:02
  • @francis-jamet, no I haven't searched for that. So all examples that you know based on axiom of choice? – Norbert Jul 02 '12 at 13:34
  • Yes, an example: if $F$ is a complement of the subspace of finite sequences, and $f_{|F} \mapsto 0$ and $f(e_n)=ne_1$. Sorry if it doesn't help. – francis-jamet Jul 02 '12 at 13:59
  • What is $f$? Is this minimal projection on $\mathrm{span}{e_1}$? – Norbert Jul 02 '12 at 14:08
  • Yes, $f$ is a projection on span ${e_1}$. – francis-jamet Jul 02 '12 at 14:14
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    @francis-jamet: It's impossible to prove that there is a discontinuous linear map without invoking some strong form of choice -- "you can't write down a discontinuous linear map defined on all of a Banach space". For instance, it follows from the axiom of determinacy that every linear map on a separable Banach space is continuous. – t.b. Jul 02 '12 at 14:45
  • @t.b. Oops, you are quite right, I was trying to use the Rademachers inside $L^1$ and then transport by finite-dimensional representability. However, one can still find copies of fin-dim $\ell^2$ inside $\ell^1$ (Dvoretzky, but I think Khintchine should work) and then take projections onto complements of these fin-dim subspaces –  Jul 02 '12 at 15:02
  • @t.b. Very interesting, thanks. – francis-jamet Jul 02 '12 at 16:05
  • If Bill Johnson did not say anything informative at mathoverflow, it is hopeless to expect an answer here, at MSE. – Moishe Kohan Apr 22 '14 at 00:40
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    Every Banach space which is not isometric to a Hilbert space contains a two-dimensional subspace which is not 1-complemented. On the other hand, every 2-dimensional Banach space embeds isometrically into $\ell_1$ (Herz–Lindenstrauss), so I think that even at the level of two-dimensional subspaces the situation can be quite difficult. Maybe you should somehow narrow yourself and ask a more specific question? – Tomasz Kania Aug 26 '14 at 14:43
  • @TomekKania, ok here is the question that bothers me for years: Let $T:\ell_1^0(M)\to X$ be a quotient map between normed spaces. For each $C>1$ we have a right inverse $S$ of $T$ of norm $<C$. Is it true that $X=\ell_1^0(N)$ for some $N$. If you can answer this question or give a counterexample I can organize it as separate MSE question and give the maximal bounty for it. Twice. – Norbert Aug 26 '14 at 18:26
  • Sorry, what is $\ell_1^0(M)$? – Tomasz Kania Aug 26 '14 at 20:24
  • @TomekKania, $c_{00}(M)$ - finitely supported sequences on $M$, $\ell_1^0(M)$ - $c_{00}(M)$ with $\ell_1$-norm. In a more fancy way my question is the following: Is it true that all projective (with respect to quotients maps) spaces in the category of normed spaces are $\ell_1^0(M)$. – Norbert Aug 26 '14 at 20:46
  • This is a result of Helemskii http://arxiv.org/pdf/1112.5750v1.pdf – Tomasz Kania Aug 26 '14 at 20:57
  • @TomekKania, this paper handles projectivity with respect to restricted class of quotient maps (they map closed unit ball onto closed unita ball), and my question is about all quotient maps – Norbert Aug 26 '14 at 21:13
  • @TomaszKania you can post this comment as an answer. I'll accept it. – Norbert Aug 06 '23 at 18:13

1 Answers1

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Without loss of generality, regard $\ell^1(S)$ for $S=\{0,1\}$.

Let $e_0=(R,0)$ and $e_1=(R,r)$ for $R$ and $r$ both nonzero.

Then $\|e_0\|_1=R$ while $\|e_1-e_0\|_1=r$.

From this we obtain a projection $P:\ell^1(S)\to\ell^1(S)$ given by $$P(e_0):=e_0\textrm{ and }P(e_1):=0,$$

whose norm is bounded from below by $R/r$.

In particular, for $R>r$ we obtain $\|P\|>1$.

freishahiri
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  • I think the op was asking about general results instead of examples. But +1 for the nice construction. – Vim Feb 05 '18 at 02:18
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    @Vim: Oh yes, true that, missread the question. But thanks for the compliment. :) Do you think I should remove or leave the answer then? – freishahiri Feb 05 '18 at 02:25
  • perhaps you can leave it till the op sees it and decides whether it's helpful or not to the question? I don't know. – Vim Feb 05 '18 at 02:28