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

Question

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?

Answer 1

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.