Assume that $X$ is a Banach space in which every Hilbertian subspace is complemented (let's say that all the projections are uniformly bounded). What can we say about $X$? It has to be Kconvex. By Maurey's extension theorem, this property holds for all spaces of type 2, so we cannot say more. But do we have something more than Kconvexity in general?

$\begingroup$ Are there any particular properties that you hope for? $\endgroup$– Tomasz KaniaOct 13, 2014 at 9:44

$\begingroup$ I'm not really sure; the question is motivated by the operator space theory and I don't know, what may turn out to be useful. $\endgroup$– Mateusz WasilewskiOct 13, 2014 at 10:36

$\begingroup$ Maybe being $\pi$Euclidean would be of use for you? See Cor. 8.4 in Projections onto Hilbertian subspaces of Banach spaces by Figiel and TomczakJaegermann. $\endgroup$– Tomasz KaniaOct 13, 2014 at 10:47

2$\begingroup$ Such a space $X$ must be of type $2\epsilon$ for all $\epsilon >0$ (since by Krivine's theorem, otherwise $\ell_p$ is finitely representable in $X$ and $L_p$ contains an uncomplemented Hilbert space). It need not be of type 2 (consider $(\sum_n \ell_{p(n)}^{k(n)})_2$ with $p(n) \uparrow 2$ and $k(n) \to \infty $ quickly). $\endgroup$– Bill JohnsonOct 13, 2014 at 11:30

1$\begingroup$ No; if $k(n)^{1/2  1/p(n)}$ stays bounded the space is isomorphic to $\ell_2$. But I was wrong about this being an example. The space does contain an uncomplemented Hilbertian subspace. Sorry about that. $\endgroup$– Bill JohnsonOct 14, 2014 at 13:34
1 Answer
There is a basic difference between "all hilbertian are complemented" and "all projections are uniformly bounded". Here is an example: Let $X$ be a separable Banach space such that every operator from $\ell_2$ into $X$ is compact. Consider $\ell_2\oplus X$ and let $Z\subset \ell_2\oplus X$ be Hilbertian. Then $P_XZ$ is compact so there exists a decomposition $Z=Z_1\oplus Z_2$ such that $\P_XZ_1\<\epsilon$ and $Z_2$ finite dimensional. Note that norms in this $\oplus$ depend only on BanachMazur distance from $Z$ to Hilbert not on $\epsilon$. It suffices to show that $Z_1$ is complemented. For $z\in Z_2$ we have $\P_{\ell_2}(z)\\geq c\z\$ so $V=:P_{\ell_2}(Z_2)$ is a closed subspace in $\ell_2$, let $Q$ be orthogonal projection from $\ell_2$ onto $V$. Let us define $S(\xi,x)=(P_{\ell_2}Z_2)^{1} Q(\xi)$. It is onto $Z_2$ and for for $z\in Z_2$ we get $\zS(z)\$ is smallso $Z_2$ is complemented. As $X$ we can take $\ell_p$ with $1\leq p<2$ but also ANY sum $(\sum_n W_n)_p$ of finite dimensional spaces $W_n$. This shows that no local conditions like type, Kconvex etc work without the uniform bound.
Sorry; it is really a comment but I got the space limit so put it here.

$\begingroup$ Yes, but it is clear from context that the proposer meant with estimateshe is using colloquial Banach space local theoryspeak. One of us should have pointed that out in the above thread. $\endgroup$ Jun 2, 2015 at 16:04