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 $\min\{h\_A,h\_B\}=h\_{A\cap B}$.



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> 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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When minimum of two supporting functionals of convex bodies is convex?
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.


