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This is a pretty easy question to ask, but haven't seen it anywhere: suppose I have some continuous path $X$ in $\mathbb{R}^n$ and I want to get the convex hull of $X$, $co(X)$.

Question: Is it enough to consider only pairwise convex combinations of points in $X$ to generate $co(X)$? in other words:

$\forall z\in co\left(X\right)\exists\lambda\in\left[0,1\right]$ and $x_{0},x_{1}\in X$ such that $z=\lambda x_{0}+(1-\lambda)x_{1}$

Also: if this is true, is it generalizable to more general topological spaces?

Thanks!

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3 Answers 3

This is true in $\mathbb R^2$ but not in higher dimensions. For example, consider a path in $\mathbb R^3$ that lies in the half-space $z\ge 0$ and touches the $xy$-plane at three non-collinear points. The convex hull contains the solid triangle spanned by these points, but pairwise convex combinations only give you three segments in that plane.

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No, it is not enough to consider convex combinations of pairs of points in the connected set. A famous example is the moment curve $(t,t^2,t^3,\dots,t^n)$ where when you take the convex hull all convex combinations of [n/2] points form a face of the convex hull. Caratheodory theorem asserts that for every $X$ in $R^n$ a point in the convex hull of X is in the convex hull of $d+1$ points from $X$. I vaguely remember that when $X$ is connected you can replace $d+1$ by $d$ but I am not sure about it.

Added later: Indeed it is an old theorem that you can replace $d+1$ with $d$ when $X$ is connected. A recent theorem of Barany and Karasev assets that if $X$ is a set in $R^d$ with the property that all projections of $X$ into a $k$ dimensional space are convex, then every point in the convex hull of $X$ is already in the convex hull of d$d+1-k$ points from $X$.

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The answer is no. For example, it's fairly easy to draw a knot $S^1 \to \mathbb R^3$ such that the convex hull is not the same thing as the set of all secants. If you want a concrete example, take a standard parametrization of a trefoil, so that the origin is the intersection of two axis of symmetry. You'll see the origin is in the convex hull, but its not on the set of secants.

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Dear Ryan, your answer came 13 seconds before mine. –  Gil Kalai Oct 6 '11 at 18:59
    
Dear Gil: and your answer is more concrete than mine. –  Ryan Budney Oct 6 '11 at 19:19
    
all three answers are nice and rather different.. i never thought about the convex hull of the standard realization of the trefoil, looks interesting. –  Gil Kalai Oct 6 '11 at 19:36
    
The geometry of the the convex hull of a smooth knot has some connections to finite-type invariants (via my work with Conant, Sinha and Scannell). It's also connected to rope-length geometry / packing type problems for knots, via work of John Sullivan, Kusner, Cantarella, Denne and that group. –  Ryan Budney Oct 6 '11 at 22:00
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And here is the Cantarella, Kuperburg, Kusner and Sullivan reference that I was thinking of, but forgot to accurately cite: front.math.ucdavis.edu/0204.5106 –  Ryan Budney Oct 9 '11 at 16:39

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