# Finite dimensional subspaces of $L^1.$

This question is motivated by my discussion (via comments) with @fedja regarding this earlier question. In any case the question is whether there is any concise characterization of finite dimensional subspaces of $L^1.$ I found some papers by our own Bill Johnson on finite dimensional subspaces of $L^p$ spaces, but they appear to (a) be for $p>1$ and (b) use a lot of language hard to understand for us troglodytes. In the meantime, the question is very concrete: I give you a centrally symmetric convex body in $\mathbb{R}^d,$ and I ask whether this is the unit ball of the sup norm of the linear combinations of some $d$ functions $f_1, \dots, f_d.$

EDIT As @Fedor points out, I asked a different question from what I intended: I had meant to ask about $L^\infty,$ but the answers to this are very interesting, so I will let it stand, and I guess will ask a different question to avoid (or increase) confusion.

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Why sup norm? Rather it should be $\|x\|=\int |f(x)| d\mu(f)$ for some measure $\mu$ on linear functionals. –  Fedor Petrov Apr 20 '12 at 19:01
Because I am asleep, that's why, you are right... –  Igor Rivin Apr 20 '12 at 19:12

I think that $K$ is such a convex body if and only if its dual $K^*$ is a zonotope, i.e., a Minkowski sum of finitely many line segments, or a limit of zonotopes. I think that every limit $Z$ of zonotopes is characterized by a finite-mass measure $\mu$ on projective space $\mathbb{R}P^{d-1}$. Then the points in $Z$ are of the form $$\int_{S^{d-1}} \vec{x} d\sigma(\vec{x}),$$ which were $\sigma$ is a measure on the sphere $S^{d-1}$ that projects to $\mu$.

You can get every zonotope as an example of $K^*$ by letting each $f_k$ be piecewise constant. Then $K^*$ is a projection of a finite-dimensional cube, and those are exactly the zonotopes. Then, every choice for $K^*$ is a limit of zonotopes, because $f_k$ has piecewise constant approximations. Actually you can make the measure $\mu$ directly by taking the pushforward of Lebesgue measure under the function $\vec{f}/||\vec{f}||$; that is a measure on the sphere but you can take its pushforward to projective space. A similar construction yields the measures $\sigma$.

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There are many characterizations of Banach spaces that embed isometrically into $L_1$. See

Wells, J. H.; Williams, L. R. Embeddings and extensions in analysis. Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 84. Springer-Verlag, New York-Heidelberg, 1975

for the classical results. Lindenstrauss and Pelczynski gave another useful one in their "Absolutely summing operators..." paper; namely, whenever $\sum |x^*(x_n)| \le \sum |x^*(y_n)|$, then $\sum \|x_n\| \le C \sum \|y_n\|$ where $C=1$. The main reason this is useful is that if you have the condition for some $C$, then the space $C$-embeds into $L_1$.

One quite remarkable fact (due to Lindenstrauss) is that every two dimensional Banach space embeds isometrically into $L_1$.

Greg, not surprisingly, gives the convex geometry way of looking at subspaces of $L_1$, while the computer scientists I know would tell you about cut metrics and try to convince you that the really interesting problem is to understand which discrete metric spaces biLipschitz embed into $L_1$.

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I assume that the distinction between $L^1$ and $\ell^1$ is intentional. Are you, in fact, saying that Greg's answer works for the latter and not for the former? –  Igor Rivin Apr 20 '12 at 19:10
The polygon case of Lindenstrauss' 2D result is easy to visualize: Every centrally symmetric polygon is a zonogon (and therefore also a dual zonogon). –  Greg Kuperberg Apr 20 '12 at 19:11
@Igor $L^1$ can only give you more, because $\ell^1$ embeds isometrically into $L^1$. I think that $\ell^1$ only gives you the duals of those limits of zonotopes whose sphere measure is pure point. –  Greg Kuperberg Apr 20 '12 at 19:14
@Greg. That would make sense... –  Igor Rivin Apr 20 '12 at 19:16
Right; the $\ell_1$ is a typo. I'll change it. –  Bill Johnson Apr 20 '12 at 19:18

Greg is right, of course, the dual must be a zonotope.

Let me mention also a direct characterization: for any vectors $x_1,\dots,x_k,y_1,\dots,y_m$ such that $\sum |f(x_i)|\geq \sum |f(y_j)|$ for any linear functional $f$, one may deduce that $\sum \|x_i\|\geq \sum \|y_j\|$. For Euclidean norm it is usual stuff in integral geometry: for proving inequality between sums of lengths one proves corresponding inequality for sums of projections to the same line. The norms, for which such trick is possible, are exactly those coming from sections of $L^1$.

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