Prove that the set of the sum of any two points on the boundary of a convex set is convex Let $C$ be a closed and bounded convex set in $\mathbb{R}^3$ and let $B = \partial C$ be its boundary. If $S = \{p+q~~|~~p,q \in B\}$, prove that $S$ itself is convex.
I'm not even sure how to get started here, even though I have a (probably incorrect) hunch that these 2 properties may be used: 1) Closure of a convex set in $\mathbb{R}^n$ is convex, and 2) if $C_1$ and $C_2$ are convex subsets of $\mathbb{R}^n$, then the set $\{x_1+x_2~~|~~x_1\in C_1, x_2 \in C_2\}$ is also convex.
Edit: The second fact is probably useless because $\partial C$ isn't necessarily convex.
(Disclaimer: This question is from the end of this video, but the video is old enough to justify asking the question here out of curiosity)
 A: It's false for unbounded convex sets. A corrected version of a counterexample in a deleted post: if $C=[-1,1]\times\Bbb R^2$ then $\partial C=\{-1,1\}\times\Bbb R^2$, so $S=\{-2,0,2\}\times\Bbb R^2$.
If $C$ is a closed bounded convex set in $\Bbb R^n$ for $n>1$ I suspect it's true, because it seems to me that $S=2C=\{2x:x\in C\}$. Convexity of $C$ shows that $S\subset 2C$; for the other direction we need this:


If $x\in C$ there exist $p,q\in\partial C$ with $x=(p+q)/2$.


If $x\in\partial C$ we can take $p=q=x$. So we can assume that $x\in C^o$:
For every unit vector $v$ choose $t(v)>0$ and $s(v)>0$ so that $x+t(v)v\in\partial C$ and $x-s(v)v\in\partial C$. If there exists $v$ such that $t(v)=s(v)$ we're done.
And if we can choose $t(v)$ and $s(v)$ to depend continuously on $v$ that follows by connectedness of the unit sphere, since if $f(v)=t(v)-s(v)$ then $f(-v)=-f(v)$, so there must exist $v$ with $f(v)=0$.


If $x$ is an interior point of $C$ and $|v|=1$ there is a unique $t>0$ such that $x+tv\in\partial C$.


Suppose otoh that $0<t_1<t_2$ and $x+t_jv\in\partial C$. The convex hull of a ball about $x$ and the point $x+t_2v$ contains a ball about $x+t_1v$; so in fact $x+t_1v$ is an interior point of $C$, contradiction.


If $x$ is an interior point of $C$ then $t(v)$ depends continuously on $v$.


Say $|v_n|=1$ and $v_n\to v$. If $t(v_n)\not\to t(v)$ then passing to a subsequence we can assume $t(v_n)\to\alpha\ne t(v)$. Now since the boundary is closed it follows that $x+\alpha v\in\partial C$, contradicting the uniqueness of $t(v)$.
QED.
