Timeline for Thin large subspaces of $\ell^N_1$
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Nov 22, 2015 at 8:47 | comment | added | ARG | many thanks for writing the details| After a night's sleep, my confusion was gone . | |
| Nov 22, 2015 at 8:43 | vote | accept | ARG | ||
| Nov 22, 2015 at 1:25 | history | edited | Mikhail Ostrovskii | CC BY-SA 3.0 |
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| Nov 22, 2015 at 0:56 | comment | added | Mikhail Ostrovskii | @Antoine See my improved answer. | |
| Nov 22, 2015 at 0:55 | history | edited | Mikhail Ostrovskii | CC BY-SA 3.0 |
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| Nov 21, 2015 at 21:35 | comment | added | ARG | well, I am utterly confused then. Because in my first paragraph, as soon as you go to larger $N$ the codimension decreases (in proportion). So putting the $\forall \epsilon$ at the beginning is not a correct reformultion. A 'No' would be [here I am putting $n = \tfrac{N}{\ln N}$] "for all $\delta$ and for arbitrarily large $N$, there is a supspace such that $\dim(V_N) \geq N - \frac{N}{\ln N}$ and $\sup \lbrace \|x\|_\infty : x \in V_N, \|x\|_1 \leq 1 \rbrace < \delta$. This is why I thought (in my first comment) that the answer had to rely on how the "constant" $C$ depends on $\alpha$. | |
| Nov 21, 2015 at 21:09 | comment | added | Mikhail Ostrovskii | OK, it shows that the answer to the question in your first paragraph is "No". It seems that you meant something else. | |
| Nov 21, 2015 at 20:49 | comment | added | ARG | Well, that does not the question I had then... nonetheless very interesting! Many thanks! | |
| Nov 21, 2015 at 19:28 | comment | added | Mikhail Ostrovskii | @Antoine I assume that you want to show that for each $\varepsilon>0$ and $\delta>0$ you can find an arbitrarily large $N$ and a subspace $V_N\subset\ell_1^N$ such that dim$(V_N)\ge (1-\varepsilon)N$ and $\sup\{||x||_\infty: x\in V_N, ||x||_1\le 1\}<\delta$. I believe that you can do this using my second and my last comments, as well as the fact that for vectors with nontrivial $||\cdot||_\infty$-norm, $\ell_1$ and $\ell_2$ norms are close to each other. | |
| Nov 21, 2015 at 18:48 | comment | added | ARG | @Milkhail:Thanks. Still I am confused... Basically, applying this result for $\alpha \in (0,1)$, can only be useful for finitely many $V_N$ since $\dim V_N = N- o(N)$ so for all $N>N_0(\alpha)$, $V_N$ is not contained in ANY space of dimension $\alpha N$. Since you can apply this result for only finitely many $V_N$, I do not see how you can take "$N$ large so that $\sqrt{N}$ kills the constant". Did I miss something obvious or could you explain your line of thought? | |
| Nov 21, 2015 at 16:25 | comment | added | Mikhail Ostrovskii | @Antoine The version of the Kashin's result needed here is: For any $\alpha\in(0,1)$ there exists $C=C(\alpha)$ such that for any $N$ there is an $[\alpha N]$-dimensional subspace $L$ of $\ell_1^N$ satisfying $$\forall x\in L\quad \frac1{\sqrt{N}}||x||_1\le||x||_2\le \frac{C}{\sqrt{N}}||x||_1.$$ | |
| Nov 21, 2015 at 14:52 | comment | added | ARG | @Milkhail, thanks again, but this reference is not to be found where I am :-(. More importantly, I am confused now with your precision: if you take the $n=c N$ Kashin section and then increase the $N$ (say to $N'$), then the subspace you have for $N'$ will not fit in those sections anymore (the codimension will be way too small)...? that is for $N'$, $V_N$ has now codimension $c' N'$ (with $c' << c$) and cannot fit inside a space of dimension $cN'$ or $(1-c)N'$... | |
| Nov 20, 2015 at 20:29 | comment | added | Mikhail Ostrovskii | @Antoine It seems that Pisier does not mention general proportion in his book; see for more general results e.g. Section 1.10.3 in Brazitikos, Silouanos; Giannopoulos, Apostolos; Valettas, Petros; Vritsiou, Beatrice-Helen Geometry of isotropic convex bodies. American Mathematical Society, Providence, RI, 2014. | |
| Nov 20, 2015 at 19:46 | comment | added | ARG | great, I still have to put my hands on the book (which might need a few days). | |
| Nov 20, 2015 at 16:56 | comment | added | Mikhail Ostrovskii | I did not think in such terms. I meant: suppose you want to show that for sufficiently large N you can pick $n=0.0001N$ such that the supremum is $<0.001$. You look at the distance-to-Euclidean constant $C$ in Kashin section of dimension $0.9999N$ and pick $N$ so large that $\sqrt{N}$ kills it (as needed). | |
| Nov 20, 2015 at 16:49 | comment | added | ARG | interesting, thanks! So you could get a counterexample if (say) $n = N^{2/3}$ but perhaps not if (say) $n= N^{1/2}$? (the choice of exponent is random and arbitrary! it was just meant to illustrate) More precisely: if there are functions $f(N)$ which are $o(N)$ so that counterexamples can be produced, what are the best function? | |
| Nov 20, 2015 at 16:43 | history | answered | Mikhail Ostrovskii | CC BY-SA 3.0 |