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 ?

  • $\begingroup$ What you're describing is a (closed) ball, not a sphere. $\endgroup$ – anomaly Jun 23 '15 at 6:16

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 $u_i = u$.


$B$ 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 $B$.

  • $\begingroup$ 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

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.