Timeline for $L^{p}$ isoperimetric inequalities on the Hamming cube
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 14, 2023 at 7:53 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tags
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| Mar 14, 2023 at 5:44 | answer | added | Paata Ivanishvili | timeline score: 2 | |
| May 19, 2017 at 13:00 | answer | added | Ryan O'Donnell | timeline score: 2 | |
| May 19, 2017 at 8:45 | comment | added | Guillaume Aubrun | For certain values of $n$ and $p<1$ you can beat both the Hamming ball and the half-cube ; for example $n=5$, $p/2=0.43$, and $A$ the set of vertices with a majority of $1$ among the first 3 coordinates. It seems reasonable to conjecture that extremizers will always be obtained as the intersection of the discrete cube with a half-space in $\mathbb{R}^n$. | |
| May 18, 2017 at 18:53 | history | edited | Paata Ivanishvili | CC BY-SA 3.0 |
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| May 18, 2017 at 18:45 | comment | added | Paata Ivanishvili | @GuillaumeAubrun It would be interesting to understand what happens for the critical exponent $p=1$. Suddenly Hamming balls stop becoming optimizer, and some kind of transition happens. | |
| May 17, 2017 at 20:05 | comment | added | Paata Ivanishvili | @GuillaumeAubrun I am not sure about $C(p)=1$ for $p\geq 1$. For $1<p\leq 2$ I can show that $C(p)\geq Z_{p/(p-1)}^{p}$ where $Z_{q}$ is the smallest positive zero of the confluent hypergeometric function $M(-q/2, 1/2, x^{2}/2)$. | |
| May 17, 2017 at 14:55 | comment | added | Anthony Quas | @GuillaumeAubrun: thanks - I removed my comment | |
| May 17, 2017 at 14:12 | comment | added | Guillaume Aubrun | @AnthonyQuas Note that raising power is $p/2$ (and not $p$) | |
| May 17, 2017 at 8:36 | comment | added | Guillaume Aubrun | Just a remark: you can ask the same question for $0 < p \leq 2$. Then as $p$ tends to zero the problem degenerates into the vertex-isopermetric inequality, whose minimizers are Hamming balls instead of half-cubes. One can check that half-cubes become better than Hamming balls when $p>p_0(n)$ with $p_0(n)<1$ and $\lim p_0(n)=1$. It is possible that $C(p)=1$ for all $p \geq 1$ ? | |
| May 16, 2017 at 22:19 | history | edited | Paata Ivanishvili | CC BY-SA 3.0 |
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| May 16, 2017 at 22:12 | history | asked | Paata Ivanishvili | CC BY-SA 3.0 |