Timeline for Exact formula or non-trivial upper bound on p-norm of $f(x)=\|x\|_2$ in $[0,1)^d$
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 26, 2021 at 17:14 | vote | accept | MikeG | ||
| Sep 26, 2021 at 16:23 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 26, 2021 at 16:18 | history | edited | LSpice | CC BY-SA 4.0 |
`\DeclareMathOperator` and `\eqref`
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| Sep 26, 2021 at 16:07 | comment | added | Giorgio Metafune | Ah, ok. Thank you | |
| Sep 26, 2021 at 16:06 | comment | added | Iosif Pinelis | @GiorgioMetafune : I have added a detail on this. | |
| Sep 26, 2021 at 16:05 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 26, 2021 at 15:58 | comment | added | Giorgio Metafune | I have not understood how you get the last inequality in your answer. | |
| Sep 26, 2021 at 15:52 | comment | added | Iosif Pinelis | @GiorgioMetafune : Good point! | |
| Sep 26, 2021 at 15:41 | comment | added | Giorgio Metafune | For $p \leq 2$ you can use a similar trick. Call $c_p$ the value of the integral so that $c_\infty=\sqrt d$, $c_2=\sqrt{d/3}$. Then $c_2 \leq c_p^{p/2}c_\infty^{1-p/2}$ gives $c_p \geq \sqrt{d} /3^{1/p}$. | |
| Sep 26, 2021 at 15:28 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 26, 2021 at 13:49 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 26, 2021 at 12:59 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 26, 2021 at 12:50 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |