Timeline for Property of $p$-norm in the $n$-simplex
Current License: CC BY-SA 4.0
14 events
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| Feb 5 at 5:29 | comment | added | Iosif Pinelis | @TomGoodwillie : Thank you for your further comment. Your argument is indeed a bit simpler. | |
| Feb 5 at 3:17 | comment | added | Tom Goodwillie | Here is an alternative argument. Let $v$ be $(0,\frac{1}{n-1},\frac{1}{n-1},\dots)$ and parametrize the line through $u$ and $v$ by writing $x(t)=u+t(v-u)$. Since $t(v-u)$ and $-t(v-u)$ have equal $p$-norm, the conjecture implies that $x(t)$ and $x(-t)$ have equal $p$-norm, so that $||x(t)||_p$ is an even function of $t$ (for small $t$). But $||x(t)||_p^p$ is not even. It is $(\frac{1}{n(n-1)})^p((n-1)^p(1-t)^p+(n-1)(n-1+t)^p)$. The third derivative of $(n-1)^p(1-t)^p+(n-1)(n-1+t)^p$ at $t=0$ is $p(p-1)(p-2)(n-1)^{p-2}(-(n-1)^2+1)$, which is not zero if $n>2$ and $p\neq 1,2$. | |
| Feb 4 at 13:44 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 4 at 12:44 | comment | added | aureliano_buendia | @IosifPinelis : it makes sense to me. Thank you very much for your answer! | |
| Feb 4 at 12:41 | vote | accept | aureliano_buendia | ||
| Feb 4 at 4:23 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 4 at 3:55 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 4 at 3:45 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 4 at 3:34 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 4 at 3:16 | comment | added | Iosif Pinelis | @TomGoodwillie : Thank you for your comment. This is now fixed. | |
| Feb 4 at 3:15 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 4 at 2:51 | comment | added | Tom Goodwillie | Well, it's also true if $p=1$, for the simple reason that $||x||_1=1$ for every $x$ in the simplex. | |
| Feb 4 at 2:46 | history | edited | Tom Goodwillie | CC BY-SA 4.0 |
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| Feb 4 at 2:39 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |