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Let $X$ be a Banach space for which there exists a constant $\beta<\infty$ such that for every finite-dimensional subspace $B$ of $X$ , $d(B,\ell_2^n)\le\beta$ (where $\dim B=n$). Then $X$ is isomorphic to a Hilbert space.

$d(B,\ell_2^n):$ Banach-Mazur distance

$\ell_2^n=(\mathbb{R}^n,||\cdot||_2)$

Any hints would be appreciated.

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marked as duplicate by Norbert, Jared, Chris Eagle, tomasz, Davide Giraudo Jul 19 '13 at 8:36

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    $\begingroup$ Do you know that it's enough to show $X$ is isomorphic to a space where the parallelogram law holds? $\endgroup$ – Kevin Carlson Jul 19 '13 at 2:52
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This is a well-known theorem; implicit in Grothendieck's Résumé; clean proof published by Joichi in 1966. Since the paper is in free access, I only describe the idea loosely.

Let $S$ be the lattice of finite-dimensional subspaces of $X$. For each $B\in S$, let $T_B$ be an isomorphism of $B$ onto a Euclidean space with indicated distortion bound. For each fixed $x\in X$, define a function $F_x:S\to [0,\infty)$ by $F_x(B)=0$ if $x\in B$, and by $F_x(B)=|T_Bx|^2$ otherwise. Then average $F_x$ over all $B$, and declare the result to be $\|x\|^2$. Finally, check that $\|\cdot\|^2$ satisfies the parallelogram law (which it does, being the average of functions that do).

The averaging process amounts to constructing an invariant mean on $S$. Some functions $F:S\to\mathbb R$ have the limit $\lim_S F (B)$ (limit makes sense since $S$ is directed by inclusion). A form of Hahn-Banach theorem extends this limit to an appropriately controlled linear functional on all bounded functions $S\to \mathbb R$.

The comparability of $\|\cdot\|$ to the original norm follows from the fact that the Hahn-Banach extension of $\lim_S F$ is pinched between $\liminf_S F$ and $\limsup_S F$.

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  • $\begingroup$ Last time I had checked (3 years ago), it was impossible to find a copy of the Résumé. Today I check again: someone (thank you!) TeXed it and made it available online (see in particular the legendary tableau p.30). Now it only remains to translate it English...But there is this book. $\endgroup$ – Julien Jul 20 '13 at 16:41
  • $\begingroup$ @julien Nice. No need to translate it into English, everyone should learn French. Of course. $\endgroup$ – 40 votes Jul 20 '13 at 16:47

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