1
$\begingroup$

Can someone please tell me the brief sketch (or any known reference) of the following results?

  1. Why $\ell_2$ is finitely representable in any infinite-dimensional Banach space?
  2. Why every Banach space is finitely representable in $c_0$?

A Banach space $Y$ is said to be finitely representable in some Banach space $X$ if for any finite dimensional subspace $F$ of $Y$ and $\varepsilon>0$ there exists an isomorphism $T:F\to X$ such that $\|T\|\|T^{-1}\|\leq 1+\varepsilon$.

$\endgroup$
  • $\begingroup$ Yes, of course. Thanks for pointing out. $\endgroup$ – Tanmoy Paul Oct 9 '18 at 19:50
  • 4
    $\begingroup$ You might want to read about Dvoretzky's theorem $\endgroup$ – Matthew Daws Oct 9 '18 at 20:01
6
$\begingroup$

1 is very hard and 2 is very easy. For 2, take an $\varepsilon$-net $(x_i)_{i=1}^m$ in the sphere of the $n$ dimensional space ($m$ depends on $n$ and $\varepsilon$), and norming functionals $f_i$'s. Now check that the map from the $n$-dimensional space into $\ell_{\infty}^m\subset c_0$ given by $x\to (f_1(x), f_2(x), \ldots, f_m(x))$ works.

For 1 there is a bit of easier proof if you don't care about the constant being $1+\varepsilon$ and if you are familiar with spreading models but still pretty involved. But standard reference for 1 is Milman-Schechtman's book

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ A very nice presentation of Dvoretzky's theorem is in the book by Albiac and Kalton. $\endgroup$ – Dirk Werner Oct 11 '18 at 7:48
4
$\begingroup$

The first one is Dvoretzky's theorem (or, more precisely, a consequence of Dvoretzky's theorem), which says not only that you can find $1+\epsilon$ isomorphic copies of $\ell_2^n$ in infinite dimensional spaces, but that you can find a $1+\epsilon$ copy of $\ell_2^n$ in any spaces whose dimension is at least $N$, where $N$ is a function of $\epsilon$ and $n$. The proof is technical, but the idea is simple. We use the idea of concentration of measure. We consider the norm $\|\cdot\|$ on the $N$-dimensional space $\mathbb{R}^N$. By John's theorem, after applying a suitable invertible linear translation, we can assume the Lipschitz constant of $\|\cdot\|$ with respect to the $\ell_2^n$ norm of $\mathbb{R}^N$ is at most $N^{1/2}$. Then one can show that, since the function $\|\cdot\|$ is concentrated around its mean (or median, depending on which version of concentration you use), there is a very small measure exceptional set of the sphere such that on the complement of the exceptional set, the norm $\|\cdot\|$ is almost exactly a fixed multiple of the $\ell_2^N$ norm. This exceptional set is small enough to be able to find an $n$-dimensional subspace which contains none of the exceptional set, and therefore the $\ell_2$ and $\|\cdot\|$ norms are almost exactly a multiple of each other on this $n$-dimensional subset.

Alternatively, if you know Krivine's theorem (which I personally find easier to understand), then you can deduce Dvoretzky's theorem. Every infinite dimensional Banach space has a basic sequence. There exists $1\leqslant p\leqslant \infty$ such that we can find $1+\epsilon$ $\ell_p^n$ spaces in the sequence (and even as blocks of the sequence). If $p=\infty$, this means $c_0$ (and by your second question, every space) is finitely representable in your space, including $\ell_2$. If $1\leqslant p<\infty$, this means $\ell_p$ is finitely representable in your space. But since $L_p$ is finitely representable in $\ell_p$, this means $L_p$ is finitely representable in your space. And $L_p$ contains an isometric copy of $\ell_2$ as the closed span of independent gaussian random variables. This is a roundabout way, but it contains a lot of interesting pieces along the way.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.