A continuous version of Carathéodory's convex hull theorem

Question

A well-known theorem of Caratheodory states that any point in the convex hull of a set $X\subset R^n$ lies in the convex hull of at most $n+1$ points of $X$. I am wondering about a version of this phenomenon where $X$ changes continuously and we want to select the points continuously.

Specifically, suppose that we have a continuous family of curves $\Gamma_t\subset R^n$, $t\in[0,1]$, such that the origin $o$ of $R^n$ is always contained in the convex hull of $\Gamma_t$. Can one find a finite number of points $x_i(t)\in\Gamma_t$ so that $t\mapsto x_i(t)$ is continuous, and $o$ is contained in the convex hull of $x_i(t)$?

By continuity here I mean that there exists a family of continuous maps $\gamma_t\colon[a,b]\to R^n$, with $\gamma_t([a,b])=\Gamma_t$, such that $t\mapsto\gamma_t$ is continuous with respect to the standard norm on $C^0([a,b],R^n)$. This is finer than the topology induced by Hausdorff distance.

Note 1: The answer is yes if the convex hull of each $\Gamma_t$ has interior points, and $o$ is one of them. Then, by Steinitz refinement of Caratheodory's theorem, for each $t$ there are $2n$ points $x_i(t)\in\Gamma_t$ which contain $o$ in the interior of their convex hull. So by continuity we can always select a finite number of points for a short time interval, and compactness of $[0,1]$ completes the argument.

Note 2: Anton Petrunin gives an example below which shows that the answer is no if the space of curves is equipped with Hausdorff topology. In particular there exists a family of intervals in the circle $S^1$ converging to $S^1$ from which one cannot select a point continuously (the intervals spin around $S^1$ faster and faster as they get longer).

Note 3: Pietro Majer gives an example below which shows that even when the points $x_i(t)$ can be chosen continuously, we cannot in general find continuous coefficients $c_i(t)$ such that $\sum c_i(t)x_i(t)=o$.

Answer 1

There could be an obstruction to continuity due to a lack of uniqueness of the representation of $o$ as convex combination. For example define $\Gamma_t$ to be, for $t\in\mathbb R$, the $\vee$-shaped simple curve in $\mathbb R^2$ with vertices, in order $$A_t:=(-1,0),\,B_t:=(0,t_-),\,C_t:=(1,t_+).$$
Then $(0,0)\in\text{co}(\Gamma_t)$ for all $t\in\mathbb R$; however for $t<0$, it is uniquely obtained as midpoint of $A_t =(-1,0) $ and $C_t =(1,0)$; for $t>0$ it is uniquely obtained as $B_t=(0,0)$, so there is no hope of a representation of $(0,0)$ as convex combination of points of $\Gamma_t$ depending continuously on $t$. (Note that for $t=0$ one has both representations, together with many others).

For a $C^1$ version, one can take as $\Gamma_t$ the graph of $[-1,1]\ni x \mapsto t^2x^2-(t_-)^2$: a similar situation occurs.

Answer 2

An example can be built on the solution of "Family of sets with no section" in my PIGTIKAL; see pages 86 and 94.

Answer 3

Here is an example which shows that the answer is no. Each $\Gamma_t$ is the graph of a smooth symmetric function on $[-1,1]$. For $t\in[0,1)$, $\Gamma_t$ lies above the $x$-axis except for a pair of bumps which descend below it. As $t\to 1$, the position of each of the bumps keeps oscillating to the right and left, faster and faster, while remaining on its respective side of the $y$-axis. At the same time the height of $\Gamma_t$ shrinks and it converges to $[-1,1]$, which is defined to be $\Gamma_1$.

Any selection of points $x_i(t)\in\Gamma_t$ to contain $o$ within their convex hull must include a point from the portion of each of the bumps on or below the $x$-axis. As the bumps constantly move side to side (by a distance bigger than the width of each bump) the first coordinates of these points are forced to oscillate indefinitely and hence cannot converge to unique limit points as $t\to 1$. So $x_i(t)$ cannot be chosen continuously.