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We define $$B_1(r,c) = \{x\in\mathbb{R}^d : \Vert{}x-c\Vert_1 \le r\}$$

Now for arbitrary constant $r \ge s > 0$, given constant $\epsilon \in (r-s,r+s)$, considering $v \in \mathbb{R}^d$ such that $\Vert{v}\Vert_1 = \epsilon$, what is the minimal volume of $B_1(r,0) \cap B_1(s,v)$ when $v$ varying?

Is there any idea or suggested reference?

I am sure that it goes to minimal when $v = (\epsilon, 0, ..., 0)$ after considering situations where $d = 2, 3$. But have no idea for strict proof. Any idea for proof?

Thank you very much.

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  • $\begingroup$ Isn't it just a straightforward application of Brunn-Minkowski? $\endgroup$
    – fedja
    May 20, 2019 at 12:45
  • $\begingroup$ Done. $\hspace{5pt}$ $\endgroup$
    – fedja
    May 20, 2019 at 15:25

1 Answer 1

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Let $v=v_1\varepsilon e_1+\dots + v_d\varepsilon e_d$ where $\sum_i |v_i|=1$ and the directions of the coordinate vectors $e_j$ are chosen so that $v_j\ge 0$. Then for any convex body $K$, $$ \sum_i v_i (B_1(s,\varepsilon e_i)\cap K)\subset B_1(s,v)\cap K $$ where the sum on the left is understood in the Minkowski sense. Thus, by BM, we have $$ |B_1(s,v)\cap K|\ge \min_j |B_1(s,\varepsilon e_j)\cap K| $$ When $K=B_1(r,0)$, all volumes in the minimum are the same and equal to your conjectured minimum. That's all.

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  • $\begingroup$ @nowhere Yes, it is. Wikipedia gives you the version for 2 bodies (in the bottom section) but the generalization to any finite number of bodies is straightforward. $\endgroup$
    – fedja
    May 20, 2019 at 22:34

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