## Why is the dimension of Gaussian variables is bounded by the dimension of the space?

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I'm looking at a probabilistic proof of a local version of Dvoretzky's theorem in Pisier's manuscript "Probabilistic Methods in the Geometry of Banach Spaces."

For each $\epsilon >0$ there is a number $\eta^\prime(\epsilon) > 0$ with the following property. Let $X$ be a Gaussian r.v. with values in a Banach space $B$ of dimension $N.$ Then $B$ contains a subspace $F$ of dimension $n = [\eta^\prime(\epsilon) d(X)]$ which is $(1+\epsilon)$-isomorphic to $\ell^n_2.$ Conversely, if $B$ contains a subspace $F$ with $F \stackrel{1+\epsilon}{\sim} \ell^n_2,$ then there is a $B$-valued Gaussian r.v. $X$ such that $d(X) \geq (1+\epsilon)^{-2}n.$

This statement uses the "dimension" $d(X)$ of a Gaussian variable, $$d(X) = \mathbb{E}\|X\|^2/\sigma(X)^2,$$ where `$$ \sigma(X)^2 = \sup \{ \mathbb{E} \xi(X)^2 \mid \|\xi\|_{B^\star} \leq 1 \} $$` is the weak variance of $X.$

For this to match the usual $n = O(\log N)$ statement of the theorem, you'd need a lower bound on $d(X)$ of order $\log N,$ and as a sanity check an upper bound of $O(N).$

Any hints or references on how to show these two bounds on $d(X)$? Pisier states that the upper bound $d(X) \leq N$ is easy to show, but I've not been able to prove even that.

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## 1 Answer

None of these estimates is trivial. The upper bound follows from John's theorem that the Banach-Mazur distance between an N dimensional normed space and an N dimensional Euclidean space is at most $\sqrt N$. The lower bound is more involved and uses the Dvoretzky-Rogers lemma. A good reference for the Gaussian approach to Dvoretzky's theorem is Pisier's book: The volume of convex bodies and Banach space geometry. For a more geometrical approach see: Milman and Schechtman: Asymptotic theory of finite-dimensional normed spaces. The theorem and it's proof is presented in other books as well.

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Thanks! I'll check out the mentioned references. – Alex Gittens Feb 11 2011 at 2:58
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