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Let $X$ be a Banach space. Suppose that every closed separable subspace $Y$ of $X$ is complemented in $X$ (i.e., there is a bounded linear projection of $X$ onto $Y$). Is $X$ necessarilly isomorphic to a Hilbert space?

Note: If $X$ is separable, or if every closed subspace $Y$ of $X$ (not necessarilly separable) is complemented in $X$, then the answer is positive, as proved by Lindenstrauss and Tzafriri (1977) (and pointed out by Harald Hanche-Olsen in the comment below).

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    $\begingroup$ I don't know, but it might be worthwhile to note (for the benefit of bystanders) that the answer is yes, if $X$ is separable, or if you drop the separability requirement on $Y$. This is due to a result by Lindenstrauss and Tzafriri mentioned here. And on a more trivial note, that you mean to ask if $X$ is isomorphic to a Hilbert space, since it is clearly not necessarily isometric to one. $\endgroup$ – Harald Hanche-Olsen Feb 11 '14 at 14:08
  • $\begingroup$ Thanks, I have corrected the question. $\endgroup$ – user124187 Feb 11 '14 at 14:24
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    $\begingroup$ I would say "yes". As pointed out by Harald, your assumption implies (by Lindenstrauss-Tzafriri) that any separable subspace $Z$ of $X$ is isomorphic to a Hilbert space. So any such $Z$ has type 2 and cotype 2. It follows that $X$ itself has type 2 and cotype 2, since these properties should be easily seen to be "separably determined" (if you write down what it means that $X$ does not have, say, type $2$, then you realize that it is possible to produce a separable subspace $Z$ which does not have type $2$). Now, apply Kwapien's theorem (!) to conclude that $X$ is isomorphic to a Hilbert space. $\endgroup$ – Etienne Feb 11 '14 at 14:58
  • $\begingroup$ Thank you. However, I am not familiar with the definition of type and cotype of a Banach space so I will have to spend some time to think about what you wrote. $\endgroup$ – user124187 Feb 11 '14 at 15:05
  • $\begingroup$ Maybe there is a simpler proof. Anyway, I can write a more detailed answer if it can help. $\endgroup$ – Etienne Feb 11 '14 at 15:15

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