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I am stuck upon the following problem. Consider the Banach-Mazur distance for $X$ ,$Y$ normed isomorphic vector spaces $$d(X,Y) = \inf \{ \| T \| \| T^{-1} \| : T \in GL(X,Y) \}$$ I would like to prove, for $\ell_p^n = (\mathbb R^n, \| \cdot \|_p )$, that $$d(\ell_1^n, \ell_2^n)\ge \sqrt n$$ without John's Theorem or volume inequalities. Can anybody lend a hand?

Thank you ever so much.

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1 Answer 1

Suppose $T:\ell_1^n\to \ell_2^n$ is an operator of norm $1$. Let $e_1,\dots,e_n$ be the standard basis of $\ell_1^n$. By our assumption $\|Te_j\|_2\le 1$ for all $j$.

There are $2^n$ vectors of the form $\sum_{j=1}^n \pm e_j$. Each of them has $\ell_1$ norm $n$. On the other hand, summing $\left\|\sum_{j=1}^n \pm Te_j \right\|_2^2$ over all choices of signs and expressing the Euclidean norm via an inner product, we get cancellation leading to
$$\sum_{\pm} \left\|\sum_{j=1}^n \pm Te_j \right\|_2^2 =2^n \sum_{j=1}^n \|Te_j \|_2^2 \le 2^n n $$ Thus, there is a vector of the form $x=\sum_{j=1}^n \pm e_j$ such that $\|Tx\|_2\le \sqrt{n}$. This proves $\|T^{-1}\|\ge \sqrt{n}$.

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I fail to see the last assert. I think we lack normalizing $x$. –  D... Oct 4 '13 at 2:44
    
@D... Line 3 says $\|x\|_1=n$. –  user98130 Oct 4 '13 at 2:45

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