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I'm stuck in two problems concerning about convex hull.

  1. Let $A,B,C \not= \emptyset$, compact sets in $\mathbb{R^n}$. Show that if $A+B=A+C$ then $\text{conv}(B)=\text{conv}(C)$
  2. Let $B\not= \emptyset$ in $\mathbb{R^n}$. Show that $n \text{conv}(B)+B=(n+1)\text{conv}(B).$ Also show that this is not true if we take some $m<n$.

I would appreciate some tip.

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  • $\begingroup$ Have you tried anything? This is not hard to show by taking, say, $b_1,b_2 \in B, \theta \in [0,1], b_3 = \theta b_1 + (1-\theta) b_2 \in conv(B)$ and argue from there. You just need to make the argument that this holds for any $b_1,b_2,b_3,\theta$ $\endgroup$ Feb 12, 2019 at 18:25
  • $\begingroup$ I don't think that 1. is true: If you take $A = \mathbb R^n$ and $B,C$ arbitrary, you get $A + B = A = A + C$ for free. But $\operatorname{conv}(B) = \operatorname{conv}(C)$ fails for arbitrary sets. $\endgroup$
    – gerw
    Feb 13, 2019 at 6:41
  • $\begingroup$ @gerw I missed one hypothesis, thanks. $\endgroup$
    – Lecter
    Feb 14, 2019 at 6:55

1 Answer 1

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This is too long for a comment, so I'm posting it as a (partial) answer. In general, it is true that $\text{conv}(A+B)=\text{conv}(A)+\text{conv}(B).$ And this is enough to prove the first part:

$t(A+B)+(1-t)(A+B) = (tA+(1-t)A) + (tB+(1-t)B) = A+B$, so the sum $A+B$ is convex, and since it also contains $A+B$, it must contain the convex hull of $A+B.$

On the other hand, if $x\in \text{conv}(A+B),$ then it is a finite sum $\sum t_i(a_i+b_i)$ for some $a_i\in A;\ b_i\in B;\ t>0;\ \sum t_i=1.$

Then, $\sum t_i(a_i+b_i)=\sum t_ia_i+\sum t_ib_i$ so $x\in \text{conv}(A)+\text{conv}(B).$

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  • $\begingroup$ Got it, thanks. $\endgroup$
    – Lecter
    Feb 12, 2019 at 19:21

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