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Jun 3, 2019 at 8:04 history edited erz CC BY-SA 4.0
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May 25, 2019 at 7:11 history edited erz CC BY-SA 4.0
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May 24, 2019 at 12:25 comment added Liviu Nicolaescu Have a look at Theorem 3.23 in this paper library.msri.org/books/Book50/files/02AT.pdf
May 24, 2019 at 12:18 comment added Liviu Nicolaescu I need to rethink the statement $\mu_n(B)=\Omega_n$.
May 24, 2019 at 12:09 comment added Liviu Nicolaescu $B$ is exactly covered by single unit ball (in the norm $\Vert-\Vert$. So its Hausdorff measure is the Euclidean measure $\Omega_n$ of the Euclidean unit ball. Identifying the span of $\{v_1,\dotsc, v_n\}$ with $\mathbb{R}^n$ we identify the paralellepiped with the unit cube $C$ in $\mathbb{R}^n$ so that $\mu_n(C)=c\bar{\mu}_n(C)=c$.
May 24, 2019 at 10:10 comment added erz @LiviuNicolaescu but how knowing $v_1,...,v_n$ find $\overline{\mu}_n(B)$? Also, it is not obvious to me, why $\mu_n(B)=\Omega_n$..
May 24, 2019 at 9:49 comment added Liviu Nicolaescu Denote by $\bar{\mu}_n$ the standard Lebesgue measure on $\mathbb{R}^n$ and by $B$ the unit ball with respect to the norm $\Vert-\Vert$ and by $\Omega_n$ the Euclidean volume of the Euclidean unit ball in $\mathbb{R}^n$. Then $\mu_n(B)=\Omega_n$ so $\mu_n=c\bar{\mu}_n$, $c=\frac{\Omega_n}{\bar{\mu}_n(B)}$.
May 24, 2019 at 6:37 history asked erz CC BY-SA 4.0