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Let $X=(X, \|\cdot\|)$ be some normed space. Let $C=[-1,1]^n$ and $H$ be a plane with equation $\sum_{i=1}^nr_i=s, 1\le s\le n.$ (Here $r_i$ are such that $Proba(r_i=1)=Proba(r_i=-1)=1/2$). The intersection $C \cap H$ is a polytope, $P(n, s)$. One can show that the plane $H$ is tiled by congruent parallelepipeds, $\Pi_j$, having vertices at $(b_1,\ldots, b_n)$ with $b_i=0, 1/m, 2/m, \ldots (m-1)/m$ or $1$, and each parallelepiped having $n-1$-dimentional volume $\sqrt n/m^{n-i}$. (Here $m$ is the factor by which we magnify $C$ to obtain a standard $m$-by qube).

I would like to get a lower bound for the following integral $$ \int_{P(n,s)}f(x)dx, \quad \mbox{where}\quad f(x)=\|\sum_{i=1}^nr_ix_i\|^q, \quad q\geq 1, x_i\in X. $$

Is it right that these parallelepipeds $\Pi_j$ would give a packing of $P(n,s )$, i.e. is it true that $$ \int_{P(n,s)}f(x)dx\geq \sum_{j=1}^{{ms}/{(m+1)}}\min_{\Pi_j}f(x)Vol(\Pi_j).$$ If so, then how to calculate this bound?

Thank you for your help.

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I don't understand much of this question, but I think you have a typo, Packing instead of paking. – yiyi Sep 9 '12 at 7:30
@ MaoYiyi: Yes, it was in the title. Thank you. – user202312 Sep 9 '12 at 7:33
No, this would not be a packing, because these parallelepiped tiled only plane $H$--not polytope $P(n,s)$. But its interesting question itself, what would be the packing for $P(n,s)$? I've asked this question here – Nick G.H. Sep 10 '12 at 20:11

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