# separable Banach space with Banach-Mazur distances to $\ell_2^n$ bounded must be isomorphic to $\ell_2$?

If $X$ is a separable infinite-dimensional Banach space and $C\in\mathbb{R}^+$ is an upper bound for the Banach-Mazur distance $d(E,\ell_2^n)$ for all $n\in\mathbb{N}$ and all $n$-dimensional $E\leq X$, why must $X$ be isomorphic to $\ell_2$?

I've been thinking along the following lines: let $\{x_n:n\in\mathbb{N}\}$ be a countable, dense, linearly independent set in $X$ and let $E_n=\langle x_1,\dots,x_n\rangle$ for all $n$. Now for each $n$ there's a linear isomorphism $T_n:E_n\rightarrow\ell_2^n$ such that $\|T_n\|\|T_n^{-1}\|\leq C$. If we can somehow put together these $T_n$ to form a linear isomorphism $T:\cup_nE_n\rightarrow\cup_n\ell_2^n$ = {finite-length sequences in $\ell^2$}, then this extends by continuity to a linear isomorphism $\bar{T}:X\rightarrow\ell_2$.

But how do we get $T$ from the $T_n$? Each $T_n$ is only defined on finitely many of the $x_i$. Maybe some sort of infinite series? Or can we choose each $T_n$ so as to agree with $T_{n-1}$, and still keep the $\|T_n\|\|T_n^{-1}\|\leq C$ property?

Many thanks for any help with this!

• This follows immediately from type/cotype considerations. From the hypotheses is immediate that $X$ must have type 2 and cotype 2, and Kwapien's Theorem implies $X$ must be isomorphic to a Hilbert space. But I am sure there must be a much more elementary explanation, whihc I cannot see right now. – Theo Mar 29 '13 at 18:21
• @Theo. Thanks. Yes, I think there is a much more elementary way of showing this. The question is from a set sheet at 3rd-year undergraduate level. – Harry Macpherson Apr 1 '13 at 14:37

Since $X$ is separable then there exist linearly independent set $\{x_k:k\in\mathbb{N}\}$ such that $X=\mathrm{cl}_X\left(\mathrm{span}\{x_k:k\in\mathbb{N}\}\right)$. Denote $E_n=\mathrm{span}\{x_k:k\in\{1,\ldots,n\}\}$ for $n\in\mathbb{N}$, then $$X=\mathrm{cl}_X(E_\infty)\quad\text{where}\quad E_\infty=\bigcup\limits_{n\in\mathbb{N}} E_n$$ Fix $n\in\mathbb{N}$, then by assumption there exsit $T_n:E_n\to\ell_2^n$ such that $\Vert T_n\Vert\Vert T_n^{-1}\Vert\leq C$. After suitable rescaling of $T_n$ we can assume that $\Vert T_n\Vert\leq 1$ and $\Vert T_n^{-1}\Vert< C$. Consider fucntion $$\langle\cdot, \cdot\rangle_{E_n}: E_n\times E_n\to\mathbb{R}:(x,y)\mapsto \langle T_n(x),T_n(y)\rangle_{\ell_2^n}$$ Since $T_n$ is an isomorphism this map is inner product, and what is more $$C^{-2}\Vert x\Vert^2\leq \langle x, x\rangle_{E_n}\leq\Vert x\Vert^2 \tag{1}$$ As the strange consequence for a fixed $x\in E_\infty$ the sequence $\{\langle x, x\rangle_{E_n}:n\in\mathbb{N}\}$ is a subset of Hausdorff compact $[0, \Vert x\Vert^2]\subset\mathbb{R}$.

On a directed set $(\mathbb{N},\leq)$ with standard ordering consider respective section filter $\mathcal{F}$ and ultrafilter $\mathcal{U}$ containing $\mathcal{F}$. Define the map $$\Vert\cdot\Vert_{E_\infty} :E_\infty\to\mathbb{R}: x\mapsto\lim\limits_{\mathcal{U}}\langle x, x\rangle_{E_n}^{1/2}$$ It is well defined becasuse limit along ultrafiler for any sequence contained in a Hausdorff compact exists and unique.

One can check that $\Vert\cdot\Vert_{E_\infty}$ is a norm satisfying parallelogram law. By Jordan von Neumann theorem we have well defined inner product $$\langle\cdot,\cdot\rangle_{E_\infty}:E_\infty\times E_\infty\to\mathbb{C}:(x,y)\mapsto\sum\limits_{k=1}^4\frac{i^k}{4}\Vert x+i^ky\Vert_{E_\infty}$$ Since $X=\mathrm{cl}_X(E_\infty)$, there is continuous extension $\langle\cdot,\cdot\rangle_X$ of $\langle\cdot,\cdot\rangle_{E_\infty}$ to the inner product on the whole $X$. From $(1)$ it follows that norm $\Vert\cdot\Vert_X$ induced by $\langle\cdot,\cdot\rangle_X$ is equivalent to the original norm of $X$. Hence identity map $$1_X:(X,\Vert\cdot\Vert)\to(X,\Vert\cdot\Vert_X):x\mapsto x$$ gives the desired isomorphism.

In the comments below Martin suggested a diagonal's method to avoid usage ultrafilters, but applicable only for separable Banach spaces. This approach can be easily generalized to arbitrary Banach spaces.

• Very nice solution. A small typo: the definition of the norm should be $\lVert x \rVert_{E_{\infty}}^{\mathbf 2} = \lim_{\mathcal{U}} \langle x,x\rangle_{E_n}$. As an (essentially equivalent) alternative to a limit along an ultrafilter you could use a Banach limit. – Martin Apr 7 '13 at 19:38
• @Martin, thank you! How do you think can we avoid usage of such non-constructive things like Banach limit or limit along ultrafilter? – Norbert Apr 7 '13 at 19:42
• I don't know whether a completely constructive solution is possible. One thing that could work would be to enumerate a dense rational subspace $(x_n)$ in $E_{\infty}$ and extract a diagonal subsequence of $\langle x_i, x_i\rangle_{E_k}$ as in Arzelà-Ascoli: first make sure that $\langle x_1, x_1 \rangle_{k_n}$ converges as $n \to \infty$. Then take a subsequence $k_n^{(2)}$ of the $k_n$ such that $\langle x_2,x_2 \rangle_{k_n^{(2)}}$ converges, etc. Then set $j_n = k_n^{(n)}$ and $\langle x_i,x_i\rangle_{j_n}$ should converge to $\lVert x_i\rVert_{E_{\infty}}^2$ and define a norm. – Martin Apr 7 '13 at 19:53
• @Norbert 1. I think you have a typo for the norm of $T_{n}^{-1}$, should be smaller than $C$.2. You don't need closure when you define $E_n$, since they are finite dimensional. – Theo Apr 7 '13 at 19:54
• @Theo 1. fixed 2. closure is necessary. For example $\ell_2$ is the closure of $\mathrm{cl}_{\ell_2}\bigcup\limits_{n\in\mathbb{N}}\ell_2^n$ – Norbert Apr 7 '13 at 19:57