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For a convex compact set $K\subset \mathbb{R}^n$ let us denote by $h_K$ its supporting functional $$h_K(\xi):=\sup_{x\in K}\langle\xi,x\rangle.$$ Thus $h_K\colon \mathbb{R}^n\to \mathbb{R}$ is a convex function.

Let $A,B\subset \mathbb{R}^n$ be two convex compact sets. It is well known (and easy to see) that if the union $A\cup B$ is convex then $\min\{h_A,h_B\}$ is convex; in fact in this case $\min\{h_A,h_B\}=h_{A\cap B}$.

Is the converse true? Namely assume that $\min\{h_A,h_B\}$ is convex. Does it follow that $A\cup B$ is convex?

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    $\begingroup$ @ChristianRemling No, I did not mix them up. I did not find an answer to my question in your link. $\endgroup$ – MKO Jun 18 '17 at 18:47
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    $\begingroup$ It seems to me that one should take maximum instead of minimum. (The supremum of union of two sets is the maximum of two suprema.) The maximum of two convex functions is always convex, and the maximum of two support functions is the support function of the convex envelope of the union. By the way, the support function of the intersection has quite a complicated description, see Remark 3 to Chapter 1.8 of Schneider's "Brunn-Minkowski theory". $\endgroup$ – Ivan Izmestiev Jun 18 '17 at 20:09
  • $\begingroup$ @IvanIzmestiev : I still think the post is correct. You are right that $\max\{h_A,h_B\}$ is always convex. But the point is that if the union of the sets is convex then $\min\{h_A,h_B\}$ is also convex because it is equal to the supporting functional $h_{A\cap B}$ of the intersection. $\endgroup$ – MKO Jun 19 '17 at 5:47
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Yes. At first, if $h=\min(h_A,h_B)$ is convex (note that it is also 1-homogeneous), it is a support function of the body $C:=\{x:\forall\xi\in \mathbb{R}^n,\langle \xi,x\rangle\leqslant h(\xi)\}$. Next, $C=A\cap B$, since the inequality $\langle \xi,x\rangle\leqslant h(\xi)$ is equivalent to a system of two inequalities $\langle \xi,x\rangle\leqslant h_A(\xi)$, $\langle \xi,x\rangle\leqslant h_B(\xi)$. Now assume that $A\cup B$ is not convex. It means that there exist $a\in A$, $b\in B$ such that the segment between $a$ and $b$ is not covered by $A\cup B$. Look at a line between $a$ and $b$, it intersects $A$ by a segment $[a_1,a_2]$ and $B$ by a segment $[b_1,b_2]$, we may suppose that $a_1<a_2<b_1<b_2$ on this line. Consider two convex compact sets: $[a_2,b_1]$ and $C=A\cap B$. They are disjoint, thus separated by a hyperplane. In other words, there exists $\xi\in \mathbb{R}^n$ such that $\sup_{x\in C} \langle \xi,x\rangle<\inf_{x\in [a_2,b_1]} \langle \xi,x\rangle$. It follows that $h(\xi)\geqslant \min(\langle \xi,a_2\rangle,\langle \xi,b_1\rangle)>h_C(\xi)$, a contradiction.

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