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Let $B \subseteq \mathbb{R}^n$ be compact. Is there some bounded set $D \subseteq \mathbb{R}^n$ with $0 \in D$ such that

$$ Conv(B) + D = B + D \quad ? $$

Here $+$ denotes the Minkowski sum and $Conv$ the convex hull.

EDIT: I should point out that I can prove this for various special cases, including when $B$ has finitely many extremal points or contains the boundary of its convex hull. Assuming the statement is true, it just seems like something that should be known (and maybe have a very simple proof..).

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  • $\begingroup$ How about $D=\operatorname{Conv}(B)$? $\endgroup$
    – user856
    Commented Jun 1, 2016 at 12:58
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    $\begingroup$ @Rahul That doesn't work. Let $B = {(1,0,0),(0,1,0),(0,0,1)}$. Then $(2/3,2/3,2/3) \in Conv(B) + Conv(B)$, but $(2/3,2/3,2/3) \notin B + Conv(B)$. $\endgroup$
    – Arno
    Commented Jun 1, 2016 at 13:44

1 Answer 1

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$D=(n+1)\cdot Conv(B)$ works.

It is trivial that $B+D\subset Conv(B)+D$; so it is sufficient to prove that $Conv(B)+D\subset B+D$. Take an arbitrary $x\in Conv(B)+D$; we prove that $x\in B+D$.

We know that $x=b+(n+1)c$ with some points $b,c\in Conv(B)$. By Charatheodory's theorem, there are some points $b_0,\ldots,b_n,c_0,\ldots,c_n\in B$ such that $b\in Conv(b_0,\ldots,b_n)$ and $c\in Conv(c_0,\ldots,c_n)$, so there are some weights $p_0,\ldots,p_n,q_0,\ldots,q_n\ge0$ such that $\sum_i p_i=\sum_i q_i=1$, $\sum_i p_i b_i = b$ and $\sum_i q_i c_i = c$. Due the symmetry, we can assume that $q_0\ge\frac1{n+1}$. Then

$$ x = b + (n+1)c = \sum_{i=0}^n p_i b_i + (n+1) \sum_{i=0}^n q_i c_i = \\ = c_0 + (n+1)\left(\sum_{i=0}^n \frac{p_i}{n+1} b_i + \big(q_0-\frac1{n+1}\big)c_0 + \sum_{i=1}^n q_i c_i \right). $$ In the last parentheses we have a convex combination of $b_0,\ldots,c_n$, so $$ \left(\sum_{i=0}^n \frac{p_i}{n+1} b_i + \big(q_0-\frac1{n+1}\big)c_0 + \sum_{i=1}^n q_i c_i\right) \in Conv(B) $$ and therefore $x\in B+(n+1)Conv(B)$.

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