Convex hull of path in $\mathbb R^2$ is the set of convex combination of 2 points of the path

Question

I fell onto this post https://mathoverflow.net/questions/77379/convex-hull-of-path-connected-sets.

The first answer is interesting and I can't find a simple argument to show that for $\mathbb R^2$ the convex hull is the set of convex combination of any two points of the path.

Any one have an idea ? I am more interesting in a proof using intermediate value theorem or fixed point theorem (or something like that) than a proof by construction (this should be fairly straight forward but a bit messy).

Answer 1

Okay: Assume $\gamma:[0,1]\to \mathbb{R}^2$ is a path. In general $$ \textrm{co}(\gamma([0,1]))=\left\{\sum_{i=1}^n s_i \gamma(t_i)|\;n\in\mathbb{N},t_i\in[0,1],\sum_{i=1}^ns_i=1\right\}, $$ so we wish to prove that

$$ \left\{\sum_{i=1}^n s_i \gamma(t_i)|\;n\in\mathbb{N},t_i\in[0,1],s_i\geq 0,\sum_{i=1}^ns_i=1\right\}=\left\{s\gamma(t_1)+(1-s)\gamma(t_2)|\;t_i\in[0,1],s\geq 0\right\} $$ Let us prove this by induction on $n$. So let $t_1<t_2<t_3$ and $s_1,s_2,s_3$ such that $\sum_{i=1}^3 s_i=1$. Observe, thus $x=\sum_{i=1}^3 s_i \gamma(t_i)$. We claim that one of the three lines $sx+(1-s)\gamma(t_i)$ must intersect $\gamma$ twice for $s\geq 0$. Clearly, they all intersect $\gamma$ once.

Assume that the lines only intersect the curve once for $i\in \{1,2\}$, then, in particular, $\gamma(t_1)-x$ and $\gamma(t_3)-x$ are not colinear, and by an affine transformation, we may assume that $x=0$ and that $\gamma(t_1)=(1,0)$ and $\gamma(t_3)=(0,1)$. Thus, the transformed $\gamma$ needs to be a curve that only intersects $(-\infty,1]\times \{0\}$ and $\{0\}\times (-\infty,1]$ at times $t_1$ and $t_3$ respectively, yet there exists $t_2\in (t_1,t_3)$ such that $0$ lies in the convex hull of $\{(1,0),(0,1),\gamma(t_2)\}$. This implies both coordinates of $\gamma(t_2)$ are negative, contradicting the assumption that $\gamma$ intersects the prescribed lines only once.

Accordingly, there exists at least one $i\in \{1,3\}$, an $s> 0$ and $t_4\in [0,1]$ such that $sx+(1-s)\gamma(t_i)=\gamma(t_4)$. This implies that $x$ lies in the convex hull of $\gamma(t_i)$ and $\gamma(t_4)$.

Thus, for general $\sum_{i=1}^n s_i \gamma(t_i),$ assume, without loss of generality, that each $s_i\neq 0,$ and apply the previous step to $\sum_{i=n-2}^n s_i \gamma(t_i)$ and the curve $t\mapsto (\sum_{i=n-2}^n s_i)\gamma(t)$ to get $\gamma(t^*_1)$ and $\gamma(t^*_2)$ and $r>0$ such that $$ \sum_{i=n-2}^n s_i \gamma(t_i)=\sum_{i=n-2}^ns_i(r \gamma(t_1^*)+(1-r)\gamma(t_2^*)), $$ and thus, $$ \sum_{i=1}^n s_i\gamma(t_i)=\sum_{i=n-2}^ns_i(r \gamma(t_1^*)+(1-r)\gamma(t_2^*))+\sum_{i=1}^{n-3}s_i\gamma(t_i) $$ This finishes the proof by induction.