Prove Convex Hull of Minkowski sum

Question

I want to prove that the following holds, where the $+$ means Minkowski sum:

$$ conv(A+B)=conv(A)+conv(B) $$

Let the convex hull of $A+B$ be $$ conv(A+B)=\sum_{j,k}\lambda_j\mu_k(a_j+b_k) $$

I don't know how to continue from here.

Answer 1

$\newcommand{\co}{\operatorname{co}}$ As already noted we have $\co (A+B) \subset \co A +\co B$. Now let $v+w \in \co A + \co B$. Write $v = \sum_i \alpha_i a_i$ and $w = \sum_j \beta_j b_j$. First we have $$ v + b_j = \sum_i \alpha_i (a_i + b_j) $$ so $v + b_j \in \co ( A +B)$. Now we have $$ v+w = \sum_j \beta_j ( v+ b_j). $$ So $v+w \in \co( \co (A+B)) = \co (A+B)$ and we are done.

Answer 2

A set $C$ is convex iff for all $t\in (0,1)$ we have $tC+(1-t)C= C$.

It follows that the Minkowski sum of convex sets $A, B$ is convex: $$t(A+B)+(1-t)(A+B) = (tA+(1-t)A) + (tB+(1-t)B) = A+B$$

Therefore, for general sets $A,B$ the sum $\operatorname{conv}(A)+\operatorname{conv}(B)$ is convex; and since it contains $A+B$, it also contains the convex hull of $A+B$. One inclusion proved.

For the opposite inclusion, pick a point in the convex hull of $A+B$. It is a convex combination of some points of $A+B$, i.e., $\sum \lambda_k (a_k+b_k)$. Since $$ \sum \lambda_k (a_k+b_k)= \sum \lambda_k a_k + \sum \lambda_k b_k \in \operatorname{conv}(A)+\operatorname{conv}(B)$$

Answer 3

A straightforward way to prove $\text{Conv}(A)+\text{Conv}(B)\subset \text{Conv}(A+B)$ Suppose $A=\{a_i\}_i$ and $B=\{b_j\}_j$. Let $x=\sum_i \alpha_ia_i\in \text{Conv}(A)$ and $y= \sum_j\beta_jb_j\in\text{Conv}(B)$, where $\alpha_i,\beta_j\ge 0$ and $\sum_i\alpha_i=\sum_j\beta_j=1$. Now notice that $x=\sum_i\alpha_ia_i=\sum_i\alpha_i(\sum_j\beta_j)a_i=\sum_{i,j}\alpha_i\beta_ja_i,$ and similarly $y=\sum_j\beta_jb_j=\sum_{j}\beta_j(\sum_i\alpha_i)b_j=\sum_{i,j}\alpha_i\beta_jb_j.$

Since $\alpha_i\beta_j\ge 0$ and $\sum_{i,j}\alpha_i\beta_j=(\sum_i\alpha_i)(\sum_j\beta_j)=1\cdot 1=1$, we have $x+y=\sum_{i,j}\alpha_i\beta_j(a_i+b_j)\in \text{Conv}(A+B),$ which implies $\text{Conv}(A)+\text{Conv}(B)\subset \text{Conv}(A+B)$.