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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.

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    $\begingroup$ If you put a 500 bounty on this its chances of getting answered will sky rocket. $\endgroup$ – Ragib Zaman Jun 10 '12 at 15:00
  • $\begingroup$ 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. $\endgroup$ – user31373 Jun 16 '12 at 18:56
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    $\begingroup$ @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. $\endgroup$ – t.b. Jul 2 '12 at 12:14
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    $\begingroup$ @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. $\endgroup$ – t.b. Jul 2 '12 at 14:45
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    $\begingroup$ 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? $\endgroup$ – Tomek Kania Aug 26 '14 at 14:43
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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 $\|P\|>1$.

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  • $\begingroup$ I think the op was asking about general results instead of examples. But +1 for the nice construction. $\endgroup$ – Vim Feb 5 '18 at 2:18
  • $\begingroup$ @Vim: Oh yes, true that, missread the question. But thanks for the compliment. :) Do you think I should remove or leave the answer then? $\endgroup$ – C-Star-W-Star Feb 5 '18 at 2:25
  • $\begingroup$ 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. $\endgroup$ – Vim Feb 5 '18 at 2:28

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