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I am reading Chapter 3: Convexity of Rudin's "Functional Analysis".

Here is the problem I'm having trouble solving (number 18):

Let $K$ be the smallest convex set in $\mathbb{ R}^3$ that contains the points $(1, 0, 1), (1, 0, -1)$, and $(\cos \phi, \sin \phi, 0)$, for $0 \le \phi \le 2 \pi $. Show that $K$ is compact but that the set of all extreme points of $K$ is not compact. Does such an example exist in $\mathbb{R}^2$?

So I need to show that this set is closed and bounded.

Could you help me with that?

Thanks.

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    $\begingroup$ What's the definition of "extreme point"? $\endgroup$ – Paul Oct 4 '14 at 8:33
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    $\begingroup$ It's $a \in K$ ( $K$ is a convex set) that is not the center of any (non-degenerate) segment in $K$, which means that if $a = \frac{x+y}{2}, \ x,y \in K$, then $a=x=y$ $\endgroup$ – Alf Taranto Oct 4 '14 at 8:41
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You can explicitly write down a parametrization of $K$, showing that it's a continuous image of a compact set. But here is a general argument.

Claim

If $A\subset \mathbb R^n$ is compact and $C$ is the smallest convex set containing $A$, then $C$ is compact.

Proof. The set $C$, also known as the convex hull of $C$ consists of all finite convex combinations of the points of $A$. But Carathéodory's theorem on convex hull, we only need the combinations of $(n+1)$ points of $A$. Let $$\Delta_n=\left\{x\in \mathbb R^{n+1}: x_i\ge 0, \ \sum x_i=1\right\}$$ be the standard $n$-dimensional simplex (a compact set). By the above, $C$ is the image of $A^{n+1}\times \Delta_{n}$ under the map $$\Phi((p_1,\dots, p_{n+1}), (x_1,\dots,x_{n+1}))=\sum_{i=1}^{n+1} x_i p_i$$ which is evidently continuous. Thus $C$ is compact.


Some further hints:

  1. The set of extreme points of $K$ contains almost all of the unit circle in the $xy$-plane, except for one point. This is why it's not closed.

  2. In two dimensions, the set of non-extreme point is open in the induced topology of $\partial K$, because open line segments are open relative to $\partial K$. This is why the set of extreme points is closed in this case.

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