Timeline for $||g_n||_{\infty} < \delta_{n-1}(g)$
Current License: CC BY-SA 4.0
5 events
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| Dec 6, 2019 at 23:22 | comment | added | Christian Remling | With $\le$, this just follows from the definitions: $\delta_{n-1}(g)$ tells us how much variation is possible for two points that agree on $\Lambda_{n-1}$. We can specifically take one point to be the one that (approximately perhaps, if the $\inf$ isn't a minimum) realizes the $\inf$ on $\Lambda_{n-1}$, and then we can't be more than $\delta_{n-1}(g)$ away from this if we leave $x$ unchanged on $\Lambda_{n-1}$, as we do when we compute the other $\inf$. | |
| Dec 6, 2019 at 23:21 | comment | added | Luísa Borsato | That's true!! I found this inequality in this paper ( iopscience.iop.org/article/10.1088/0951-7715/24/10/014/meta ), if you want to take a look. | |
| Dec 6, 2019 at 23:17 | comment | added | Christian Remling | Strict inequality is clearly not possible in general since we can have $\delta_n(g)=0$. | |
| Dec 6, 2019 at 22:24 | history | edited | YCor |
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| Dec 6, 2019 at 21:30 | history | asked | Luísa Borsato | CC BY-SA 4.0 |