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 K-convex. 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 K-convexity in general?
$\begingroup$
$\endgroup$
9
asked Oct 13, 2014 at 9:29
-
$\begingroup$ Are there any particular properties that you hope for? $\endgroup$– Tomasz KaniaCommented Oct 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 WasilewskiCommented Oct 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 Tomczak-Jaegermann. $\endgroup$– Tomasz KaniaCommented Oct 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 JohnsonCommented Oct 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 JohnsonCommented Oct 14, 2014 at 13:34
|
Show 4 more comments
1 Answer
$\begingroup$
$\endgroup$
1
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 $P_X|Z$ is compact so there exists a decomposition $Z=Z_1\oplus Z_2$ such that $\|P_X|Z_1\|<\epsilon$ and $Z_2$ finite dimensional. Note that norms in this $\oplus$ depend only on Banach-Mazur 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 $\|z-S(z)\|$ is small--so $Z_2$ 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, K-convex 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 estimates--he is using colloquial Banach space local theory-speak. One of us should have pointed that out in the above thread. $\endgroup$ Commented Jun 2, 2015 at 16:04