The Minkowski sum of two convex sets is convex

convex-analysis

Solution 1:

For $e,f\in C$, let $a,b\in A$ and $c,d\in B$ s.t. $e=a+c$ and $f=b+d$,

$$te+(1-t)f=t(a+c)+(1-t)(b+d) = (ta+(1-t)b) + (tc+(1-t)d)\in A+B=C$$ For all $t\in [0,1]$. $\square$

Solution 2:

If $c_1=a_1+b_1$ and $c_2=a_2+b_2$ belong to $C$ (with $a_1,a_2\in A$ and $b_1,b_2\in B$), then for $\alpha\in [0,1]$: $$\alpha c_1+(1-\alpha) c_2=\left(\alpha a_1+(1-\alpha) a_2\right)+ \left(\alpha b_1+(1-\alpha) b_2\right)$$ What may we conclude?

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