# Is Minkowski sum of boundary convex again?

Consider a closed, bounded and convex set $$C \subset \mathbb{R}^{2}$$ and denote its boundary with $$\partial C$$. It is very well-known 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?

• @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. – Guillaume Aubrun May 7 at 19:37
• 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 $11/10$. – Deane Yang May 7 at 20:18
• @DeaneYang The question is about the sum of $\partial C$ with itself. – M. Winter May 7 at 20:20
• @M.Winter, oy. Thanks. – Deane Yang May 7 at 20:24

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 $$f :S^{n-1} \to \mathbf{R}$$ 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 Borsuk-Ulam, if you want), which gives the desired property.

• 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 $-u$). – M. Winter May 7 at 20:19
• @GuillaumeAubrun Although it seems intuitively correct, can you give an argument why f is continuous? – Hermann May 8 at 0:02
• 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 $S^{n-1}$. – Guillaume Aubrun May 8 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 Brunn-Minkowski 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.

• 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. – Guillaume Aubrun May 7 at 20:02
• @GuillaumeAubrun Fair point. – Igor Rivin May 7 at 20:06
• @GuillaumeAubrun but see the edit. – Igor Rivin May 7 at 20:10
• @GuillaumeAubrun And thanks for the downvote! – Igor Rivin May 7 at 21:09
• 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. – Guillaume Aubrun May 7 at 21:30