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 $\|T\|\|T^{-1}\|\leq 1+\varepsilon$.

  • $\begingroup$ Yes, of course. Thanks for pointing out. $\endgroup$ Oct 9, 2018 at 19:50
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    $\begingroup$ You might want to read about Dvoretzky's theorem $\endgroup$ Oct 9, 2018 at 20:01

2 Answers 2


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

  • $\begingroup$ A very nice presentation of Dvoretzky's theorem is in the book by Albiac and Kalton. $\endgroup$ Oct 11, 2018 at 7:48

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.


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