8
$\begingroup$

Is there an elementary proof of this Banach space fact?

If the Banach space $V$ is linearly isomorphic to $l^1$, then it does not isometrically contain euclidean spaces of arbitrarily large finite dimension, i.e., a copy of $l^2_n$ for all $n$.

Failing that, good references on the subject?

Improve this question
$\endgroup$
6
  • $\begingroup$ (Maybe I should have specified: real scalars.) $\endgroup$
    – Nik Weaver
    Commented Apr 18, 2016 at 12:21
  • 2
    $\begingroup$ @BenMcKay: read the question more carefully. $V$ is not isometrically isomorphic to $l^1$, only linearly isomorphic. $\endgroup$
    – Nik Weaver
    Commented Apr 18, 2016 at 13:34
  • 4
    $\begingroup$ It is not a fact, Nik: Given any $\varepsilon>0$ and any infinite dimensional Banach space $X$, there is a norm $|\cdot |$ on $X$ that is equivalent to the original norm and, for all $n$, $\ell_2^n$ embeds isometrically into $(X,|\cdot |)$. This is a consequence of Dvoretzky's Theorem. I think you can even make the $|\cdot |$ norm $1+\epsilon$-equivalent to the original norm, thought that would take a bit more effort. $\endgroup$
    – Bill Johnson
    Commented Apr 18, 2016 at 13:43
  • $\begingroup$ Thank you, Bill, and everyone else for setting me straight. $\endgroup$
    – Nik Weaver
    Commented Apr 18, 2016 at 13:44
  • $\begingroup$ "linearly isomorphic"??? I guess you want your map to be not only linear, but continuous??? $\endgroup$
    – Gerald Edgar
    Commented Apr 18, 2016 at 14:24

2 Answers 2

Reset to default
5
$\begingroup$

Is it true? $l^1$ is a sum of finite-dimensional $l_n^1$ over $n=1,2,\dots$. In summands you have almost spherical sections of large dimensions by Dvoretzky theorem, this allows to change norm a bit so that unit balls in summands contain large spherical sections.

Improve this answer
$\endgroup$
4
$\begingroup$

There does not exist an elementary proof. Sadly, there is no proof at all, because the fact is false. Here is a construction of an $\ell^1$-space containing every $\ell^2_n$.

Consider the normalized Gaussian measure $d\mu=\pi^{-1/2}e^{-t^2}dt$. Form its infinite dimensional tensor product $d\mu_\infty$ over ${\mathbb R}^{\mathbb N}$. Now let $E$ denote the vector space of linear functions $$f_a:x\longmapsto a\cdot x,\qquad x\in{\mathbb R}^{\mathbb N},$$ where $a$ has finite support. $E$ is contained in $L^1(d\mu_\infty)$. One sees easily, from the rotational invariance of $d\mu$, that $$\|f_a\|_1\left(=\int|f_a(x)|d\mu_\infty(x)\right)=C\|a\|_2$$ where $$C=\int_{\mathbb R}|t|d\mu(t).$$ Hence $L^1(d\mu_\infty)$ contains every $\ell^2_n$.

Nota. This argument is used to prove that Euclidian spaces satisfy the Hlawka Inequality. The latter is almost trivial in an $\ell^1$ space.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thank you, I guess I was confused. I'm accepting Fedor's answer because it was first. $\endgroup$
    – Nik Weaver
    Commented Apr 18, 2016 at 13:43
  • 2
    $\begingroup$ If $\mu$ is not purely atomic, then $L_1(\mu)$ contains a subspace isometrically isomorphic to $\ell_2$. Nik's question was for the case where $\mu$ is purely atomic. $\endgroup$
    – Bill Johnson
    Commented Apr 18, 2016 at 13:57
  • $\begingroup$ You're assuming that there is no proof of $0=1$. Gödel would like to have a word with you. ☺ $\endgroup$
    – Gro-Tsen
    Commented Apr 18, 2016 at 17:48
  • $\begingroup$ @Nik. You can still change your choice. I had to do that once for a question of mines. Fedor's answer is perfectly right and it enlightens mine, although it does not give the definitive picture. $\endgroup$
    – Denis Serre
    Commented Apr 18, 2016 at 19:14
  • 1
    $\begingroup$ Both $L_1$ and $\ell_1$ are paved by $\ell_1^n$-s, Nik. I think Denis was making the point that Dvoretzky's Theorem for $L_1$ spaces goes way back. Another way to see that is to observe that $L_1^{2^n}$ contains a length $n$ sequence of IID Bernoulli random variables and use Berry-Essen (or even just CLT in the right form) to get random variables that approximate IID N(0,1) random variables. It is basically obvious that IID N(0,1) random variables are, in the $L_1$ norm, isometrically equivalent to an orthogonal sequence in a Hilbert space. $\endgroup$
    – Bill Johnson
    Commented Apr 20, 2016 at 14:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .