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Does every contractively complemented subspace of $c_0(I)$ is isometric to $c_0(J)$ for some $J\subseteq I$?

May be someone has a counterexample?

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  • $\begingroup$ "Has" is correct; "have" is not. (So I have rolled back your change) $\endgroup$
    – Yemon Choi
    Commented Nov 18, 2014 at 18:36
  • $\begingroup$ Why did not not correct the other three mistakes, Yemon? :) $\endgroup$ Commented Nov 19, 2014 at 1:57

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Yes, it is true, but I don't know where it is proved in the literature. The reason I say yes is that what you want is a corollary of a unpublished almost isometric theorem that Zippin proved in the 1970s! Here is either an outline of a proof or complete nonsense: Let $P$ be a contractive projection on $c_0(I)$. Then the range of $P^*$ is a sublattice of $\ell_1(I)$ (see H. Elton Lacey's book, Springer Grundlehren Band 208, The isometric theory of the classical Banach spaces) and hence is the norm closed span of a set $x_\alpha$ of disjointly supported unit vectors. For each $\alpha$, compose $P^*$ with the restriction mapping $R_\alpha$ onto the support of $x_\alpha$ to get a weak$^*$ continuous contractive projection onto the span of $x_\alpha$. So there is $y_\alpha$ in $c_0(I)$ s.t. $R_\alpha P^* = x_\alpha\otimes y_\alpha$. Argue that $(y_\alpha)_\alpha$ is 1-equivalent to the $c_0(J)$ basis for some $J$ and that the range of $P$ is the closed span of $(y_\alpha)_\alpha$

Notice that for this to happen each $x_\alpha$ must achieve its norm and thus has finite support. Interestingly, the $y_\alpha$ need not have finite support.

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  • $\begingroup$ Thank you sir! Could post in comments what was that theorem by Zippin? $\endgroup$
    – Norbert
    Commented Jul 17, 2013 at 19:06
  • $\begingroup$ A script $L_{\infty,\epsilon}$ subspace of $c_0$ is $f(\epsilon)$ isomorphic to $c_0$, where $f(\epsilon) \to 1$ as $\epsilon \to 0$. $\endgroup$ Commented Jul 17, 2013 at 19:14
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As you may know already, every contractively complemented subspace of $\ell_1(I)$ is isometric to $\ell_1(J)$ for some $J \subset I$ --- see for example Semadeni, Banach Spaces of Continuous Functions, 1971, p. 488. This is just the dual of your desired proposition. If $X$ is a contractively complemented subspace of $c_0(I)$ by a projection $P$, then the adjoint $P^*$ (as in Bill's answer above) is an isometric embedding of $X^*$ into $\ell_1(I)$ such that $P^*(X)$ is contractively complemented by the restriction operator, so that $P^*(X)$ is isometric to some $\ell_1(J)$. Can you then argue by w*-continuity of the isometry that $X$ must be isometric to $c_0(J)$?

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  • $\begingroup$ From your words it seems like a standard trick but I don't know it $\endgroup$
    – Norbert
    Commented Sep 11, 2013 at 6:13
  • $\begingroup$ Knowledge sometimes degrades over time. Something "well known" 44 years ago may degrade to "known", then to "forgotten". $\endgroup$ Commented Sep 11, 2013 at 16:58
  • $\begingroup$ Have you confirmed the validity of the last sentence in my answer, stated as a question? $\endgroup$ Commented Sep 11, 2013 at 17:07
  • $\begingroup$ No I havn't, I really have no idea how to prove the last statement. It seems you need to know more about structure of $X^*$. But in this case, imho, one had to repeat Bill's argument. $\endgroup$
    – Norbert
    Commented Sep 11, 2013 at 17:29

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