Consider a closed, bounded and convex set $C \subset \mathbb{R}^{2}$ and denote its boundary with $\partial C$. It is very wellknown that the Minkowski sum of two convex sets is convex again. What about the Minkowski sum of its boundary? Is the Minkowski sum $\partial C + \partial C$ again a convex set and how can one prove that? Does this property hold in other dimensions?

$\begingroup$ @IgorBelegradek No, $\partial C +\partial C = 2C$ in that case. I believe this equality is true in general (equivalently, any point in a convex body is the midpoint of boundary points) and should be provable by a topological argument. $\endgroup$– Guillaume AubrunMay 7, 2020 at 19:37

$\begingroup$ Take $C$ to be the unit disk and $D$ to be the disk of radius, say, $1/100$. The sum of the convex sets is the disk of radius $11/10$ but the sum of the boundaries is the annulus with inner radius $9/10$ and outer radius $11/10$. $\endgroup$– Deane YangMay 7, 2020 at 20:18

1$\begingroup$ @DeaneYang The question is about the sum of $\partial C$ with itself. $\endgroup$– M. WinterMay 7, 2020 at 20:20

$\begingroup$ @M.Winter, oy. Thanks. $\endgroup$– Deane YangMay 7, 2020 at 20:24
2 Answers
Yes, $\partial C + \partial C$ is convex since it equals $2C$. Equivalently, every point in $z \in C$ is a midpoint of two boundary points. This is obvious if $z \in \partial C$. Otherwise, let $f :S^{n1} \to \mathbf{R}$ be the continuous function which sends $u$ to the length of the segment going from $z$ to $\partial C$ in direction $u$. Since $n > 1$, this function takes equal values at a pair of antipodal points (a very simple corollary to BorsukUlam, if you want), which gives the desired property.

3$\begingroup$ If you consider segments going from $z$ to $\partial C$ in directions $u$ and $u$, and you compare their lengths, then I think all you need here is the intermediate value theorem to find that they will be equal in length at some point (by continuously moving $u$ to $u$). $\endgroup$ May 7, 2020 at 20:19

$\begingroup$ @GuillaumeAubrun Although it seems intuitively correct, can you give an argument why f is continuous? $\endgroup$– HermannMay 8, 2020 at 0:02

1$\begingroup$ This follows e.g. from the fact that the gauge function of a convex body is continuous. The gauge function $g_K$ of a convex body $K$ is defined on $\mathbf{R}^n$ by $g_K(x) = \inf \{ t \geq 0 \, : \, x \in tK \}$ ; it is convex hence continuous. When $z=0$ (which you can assume), our $f$ is the restriction of $1/g_C$ to $S^{n1}$. $\endgroup$ May 8, 2020 at 7:12
Convex hull of Minkowski sum is the Minkowski sum of convex hulls. The proof is Theorem 1.1.2 in
Schneider, Rolf, Convex bodies: the BrunnMinkowski theory, Encyclopedia of Mathematics and Its Applications. 44. Cambridge: Cambridge University Press. xiii, 490 p. (1993). ZBL0798.52001.
...But the answer to the question is: NO. See the picture in https://www.geometrie.tuwien.ac.at/peternell/ms_paper_v1.pdf (Peternell, Minkowski sum of boundary surfaces...) Can't find a citation.

1$\begingroup$ I don't see how this answers the question. Also, the desired property is false in dimension $1$, so this restriction should enter at some point. $\endgroup$ May 7, 2020 at 20:02



$\begingroup$ @GuillaumeAubrun And thanks for the downvote! $\endgroup$ May 7, 2020 at 21:09

1$\begingroup$ I did note downvote your answer and I'm sorry that you believe this! I think the picture in the paper you mention addresses a different question, namely whether $\partial C + \partial D$ is convex. $\endgroup$ May 7, 2020 at 21:30