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EDIT: Stuck on a part of a problem for a Linear programming course. I want to show the unit disc is not a convex polytope and my strategy is to show that it has infinitely many extreme points.

How would you show that every point along the boundary of the unit disc

$$D:=\{x \in R^2| x_1^2+x_2^2 \leq 1 \}$$

is an extreme point perhaps using the definition:

$v$ (a point in $S$) is an extreme point of $S$ (a convex set) if there are no distinct points $x_1$ and $x_2 \in S$ such that:

$$v = \lambda x_1+ (1-\lambda)x_2$$ where $$0 \lt \lambda \lt 1$$

EDIT: What I've tried so far:

is trying to write a point $v$ along the boundary as an interior point with two distinct points in the set and there will be a contradiction showing one must lie outside the set.

$v = (x_0, y_0)$ As $v$ lies on the boundary: $x_0^2+y_0^2=1$.

Let $t, u \in D$ be two distinct elements. $t = (x_1, y_1), u = (x_2, y_2)$

Then if $v = \lambda t + (1-\lambda)u$ , where $0 \lt \lambda \lt 1$

or

$(x_0,y_0) = \lambda (x_1, y_1) + (1-\lambda) (x_2, y_2)$ where $0 \lt \lambda \lt 1$

$\implies [\lambda x_1 + (1-\lambda) x_2]^2 + [\lambda y_1 + (1-\lambda) y_2]^2 = 1$

Then I'm stuck on showing how this isn't possible if $0 \lt \lambda \lt 1$ proving that $v$ is in fact an extreme point and that this is the case for all boundary points and there are infinitely many of them.

Am I on the right track? What am I doing incorrectly

Mathstatsstudent
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  • What have you tried so far? –  Oct 05 '19 at 01:21
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    Take any two distinct points, calculate, and see what you get. In particular, on what condition does $v$ lie on the boundary? – incertia Oct 05 '19 at 01:42
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    Hint: if $(1,0)$ is an extreme point, then so is every point on the boundary of $D$ (why?) so it suffices to show this for $(1,0).$ Another way: any point on $D$ strictly between two other points has norm strictly less than at least one of the two points. – Matematleta Oct 05 '19 at 02:06
  • $v$ lies on the boundary when $x_1^2 + x_2^2 = 1$. I think if I take two distinct points it will follow at least one must lie outside the set $D$ for them to equal $v$ using the definition above. Just having a little bit of trouble getting the algebra to work... seems more complicated than it should be – Mathstatsstudent Oct 05 '19 at 03:57
  • How about a geometric reasoning? $\lambda x_1 + (1 - \lambda)x_2$ runs through the interval $[x_1,x_2]$ as $\lambda$ runs through $[0,1]$. $0<\lambda <1$ is the same as saying that $v$ is in the $\textbf{interior}$ of some interval, which is not possible if $v$ is on the disk. – user 1987 Oct 05 '19 at 15:37

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