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How do I show that a closed ball in $\mathbb R^3$ has an infinite number of extreme points ?

(Closed ball is written as $S = \{(x,y,z) \in \mathbb R^3 | \sqrt {x^2 + y^2 + z^2} \le R \}$)

I know that $z$ is an extreme points of a convex subset $C$ if and only if $z \in \text{conv}(\{x,y\}) \Rightarrow z = x \lor z = y$.

Geometrically, I do verify that this property hold.

Also, $z$ is an extreme point if it's a face of dimension $0$.

Can someone help me proving this rigorously ?

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  • $\begingroup$ What you're describing is a (closed) ball, not a sphere. $\endgroup$ – anomaly Jun 23 '15 at 6:16
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Rather denot by $B = \{(x,y,z) \ | \ x^2 + y^2 + z^2 \le R^2\}$ your set, that is $\{ u \ | \ ||u||\le R\}$

Recall the Cauchy inequality $$||u||^2 ||v||^2 \ge \langle u, v \rangle ^2$$ with equality if and only if the vectors $u$, $u$ are proportional.

Assume now $||u|| = R$, then for every $v \in B$ we have $$|\langle u,v\rangle | \le ||u|| \cdot ||v|| = R \cdot ||v|| \le R^2$$ We have the equality $$|\langle u,v\rangle | = R^2 $$ if and only if $v = \pm u$. We conclude that if $||u|| = R$ then for every $v$, $||v|| \le R$ we have: $$\langle u,v \rangle \le R^2$$ with equality if and only if $v=u$.

Moreover, if $v \in \mathbb{R}^3$ so that $\langle u, v\rangle \le R^2$ for every $u \in S$ then necessarily $||v|| \le R$. Indeed, apply this with $u = \frac{R}{||v||} \cdot v$.

Now we see that $$B = \cap_{u \in S} \{ v \ | \ \langle u,v\rangle \le R^2 \}$$

an intersection of hyperplanes, hence a convex set ( this of course can be proved directly).

Moreover, every point $u \in S$ is the intersection of the ball $B$ with the hyperplane $H_u \colon = \{ v \ | \ \langle u,v\rangle =R^2 \}$ ( $u$ is an exposed point). It follows that the point $u$ is extreme. Indeed, if $u = \lambda_1 v_1 + \lambda_2 v_2$ then $\langle u, v_i \rangle \le R^2$ so we get $\langle u, \sum \lambda_i v_i \rangle \le R^2$. However, we do have equality, so we need to have equality $\langle u, u_i \rangle = R^2$ whenever $\lambda_i \ne 0$, but that implies $u_i = u$.

$\bf{Added:}$

$B$ has only $0$-dimensional faces and $B$ itself. Why? A face is an extreme convex subset. That is, if a point $u \in F$ has a decomposition $u = \sum \lambda_i u_i$, $\lambda_i >0$, $\sum \lambda_i =1$, then $u_i \in F$. Now, take a face that contains two points $u_1$, $u_2$. The full open segment $(u_1, u_2)$ is in the interior of $B$. Take the point $1/2( u_1 + u_2)$. This is in F. For any other point $u \in B$, there exists a segment with one end at $u$ and containing $1/2( u_1 + u_2)$ inside ( because $1/2( u_1 + u_2)$ is in the interior of $B$). But that implies $u\in F$.

What we have used is $B$ is strictly convex, that is, whenever $u_1$, $u_2$ are two distinct points on the boundary $\partial B$, the open segment $(u_1, u_2)$ is in the interior of $B$.

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  • $\begingroup$ Can you also prove that $S $ has no faces of dimension one or two ? $\endgroup$ – Shuzheng Jun 23 '15 at 9:10
  • $\begingroup$ Yea, just added some $\endgroup$ – Orest Bucicovschi Jun 23 '15 at 9:27
  • $\begingroup$ So you get a contradiction that $u \in F$ ? How do you know you can take some point $u \notin F$ ? $\endgroup$ – Shuzheng Jun 23 '15 at 11:44
  • $\begingroup$ @user111854: It shows that any face that contains more than one point of $S$ is the whole $B$. $\endgroup$ – Orest Bucicovschi Jun 23 '15 at 19:21
  • $\begingroup$ Can you write this segment with $u $ explicit and prove it is in $B $ ? $\endgroup$ – Shuzheng Jun 24 '15 at 5:30
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Points on the surface (which form an $(n-1)$-sphere) of an $n$-disc, $D^n:=\left\{v \in \mathbb{R}^n\,|\,\|v\|_2\leq R\right\}$, where $R>0$ and $n \in \mathbb{N}$, are all the extreme points of $D^n$.

Hint: See Minkowski's Inequality (http://www.imomath.com/index.php?options=595&lmm=0).

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