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This is the question I had meant to ask when I asked this question.: Is there a concise characterization of finite dimensional subspaces of $L^\infty?$ (that's what the discussion with @fedja was really about...)

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That's much easier and more standard than finite-dimensional subspaces of $L^1$. The answer is all norms in finite dimensions, or in the unit ball picture, all centrally symmetric convex bodies. Every polytope is a slice of an $n$-cube, so clearly you get all of those. But then the Banach-Alaoglu theorem in this case lets you take limits; the space of $d$-dimensional subspaces of $L^\infty$ is compact. The same argument works for $\ell^\infty$, which embeds in $L^\infty$.

In fact in the case of $\ell^\infty$ you can simply explicitly make every centrally symmetric convex body in $d$ dimensions as an intersection of countably many antipodal pairs of half spaces, and take the corresponding embedding in $\ell^\infty$.

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You mean "every centrally-symmetric polytope", I assume. Is that obvious? –  Igor Rivin Apr 20 '12 at 19:29
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If a centrally symmetric $d$-polytope has $2n$ facets, then it is a slice of an $n$-cube, which also has $2n$ facets. You can construct the linear map directly from the hyperplanes of opposite pairs of facets. The word "obvious" doesn't do much for me, but let's say that it's an exercise. –  Greg Kuperberg Apr 20 '12 at 19:35
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Greg said it right. Every centrally-symmetric polytope embeds isometrically into the much smaller space $c_0$; in fact, a finite dimensional Banach space embeds isometrically into $c_0$ if and only if its unit ball is a polytope. –  Bill Johnson Apr 20 '12 at 19:36
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Sure. All finite dimensional Banach spaces.

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Perhaps you meant to ask something slightly different? –  Bill Johnson Apr 20 '12 at 19:24
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That's what I thought. Why is that obvious? –  Igor Rivin Apr 20 '12 at 19:24
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Maybe I did (as you see I am not very with it;) In the alluded to discussion, the OP had asked for a family of functions with prescribed $2-$ and $1-$ norms, and with sup norm bounded by $1.$ As pointed out by @fedya, this last condition is the hard one (otherwise you construct a geodesic on the sphere in three dimensions and you are done), but the set of linear combinations of three given functions with sup norm $\leq 1$ appears to be an arbitrary convex centrally symmetric set (as you seem to confirm...) so... –  Igor Rivin Apr 20 '12 at 19:27
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Every separable Banach space is isometrically isomorphic to a quotient of $\ell_1$, hence every separable reflexive Banach space embeds isometrically into $\ell_\infty$, which in turn embeds isometrically into $L_\infty$. –  Bill Johnson Apr 20 '12 at 19:30
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Or: every separable Banach space embeds isometrically into $C[0,1]$, which clearly embeds isometrically into $L_\infty$. –  Bill Johnson Apr 20 '12 at 19:32
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