# A continuous version of Carathéodory's convex hull theorem

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 $$\sum c_i(t)x_i(t)=o$$.

## 3 Answers

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.

• (notation: $t_+:=\max\{0,t\}$, $t_-:=\max\{0,-t\}$) Feb 29 at 0:20
• I don't think that this example works because one could always include the point $B_t$ (I am not assuming that the number of points is minimal). Feb 29 at 1:13
• Sorry I misunderstood the question. (Here the coefficients of the convex combination of $A_t,B_t,C_t$ to express $0$ are not continuous). Feb 29 at 4:46

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

• In this example the family of curves appear to be continuous only with respect to the Hausdorff distance, since one of the curves is actually closed. In the problem I had in mind, the family of curves are continuous with respect to the C^1 norm as maps from an interval to R^n. So I do not see that the example you mention works in the sense that I intended. I will make the notion of continuity more explicit in the question. Feb 29 at 17:28

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.

• Could you clarify what you mean by this: "As the bumps constantly move side to side (by a distance bigger than the width of each bump)..."? Mar 1 at 17:17
• If the bumps only oscillated a little bit, we could still choose the points x_i(t) so that their first coordinate would remain constant and thus would converge to a point on the x-axis as t-->1. One can avoid this by moving the bumps sufficiently far and back so that their support on the x-axis have empty intersection. Mar 1 at 17:30