The intersection of finite number of convex hulls is a convex hull.

Question

What is the easiest way to prove that the intersection of finite number of convex hulls is a convex hull?

It seems to be an easy statement to prove, but still I can not get even the idea of proving. So, I would be grateful even for an idea behind the prove.

Edit: Information pulled from comments: We are talking about in the plane, and a "convex hull" here is a convex polygon - that is, the convex hull of a finite set of points.

Answer 1

Lemma: In the plane, a set $X$ is the convex hull of a finite set of points if and only if $X$ is convex and it is either a single point, a line segment, or can be written as the union of a finite number of triangles (where a triangle includes its boundary and interior.)

The we need to prove:

1. The intersection of two convex sets is convex.
2. The intersection of two triangles is a convex hull (where an empty set is considered the convex hull on an empty set.)
3. The intersection of a line segment and a triangle is either a point, a line segment, or empty.
4. The intersection of a line segment and a line segment is either a point, a line segment, or empty.

From this, you get that if $X$ and $Y$ are both convex hulls of finite sets, then $X\cap Y$ is a convex hull of a finite set.

Then you get your full result by induction.

Each of these parts are relatively easy to show, with (2) requiring the most details.

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There are two possible definitions of "convex hull" here. There is a "usual definition," but there are reasons in this question to think that your question is asking about another definition.

Let $X\subseteq \mathbb R^n$. The the "convex hull of $X$" is the smallest convex set that contains $X$. In particular, this can be written as the set of points:

$$H(X)=\left\{x\in\mathbb R^n\mid x=t_1x_1+t_2x_2+\cdots+t_nx_n,\,x_i\in X,\,t_i\geq 0, \sum t_i=1\right\}$$

Now, one might define "a convex hull" to be any $H(X)$, but (1) that's uniteresting, because then "a convex hull" is the same as a convex set - if $Y$ is a convex set, then $H(Y)=Y$. Also, with this definition of "convex hull," you don't need the "finite" part of your question.

On the other hand, one might define "a convex hull" as any $H(X)$ where $X$ is finite. In that definition, the nature of $H(X)$ is that it's boundaries are all 'flat.' But it is also tricky to prove that a finite intersection of such hulls is still such a hull.

From your comment above, particularly the reference to polygons, it does seem like you mean this second type: $H(X)$ for some finite set $X$.