# Compact convex set representation $K = \operatorname{cl} \operatorname{conv} \operatorname{exp} K$

I am reading a proof from https://www.fmf.uni-lj.si/~lavric/hug&weil.pdf (Theorem 1.5.4, page 39-40) and I don't understand few things.

1. At the beginning they mention a farthest point $$y_{x}$$:

Since $$K$$ is compact, for each $$x \in \mathbb{R}^{n}$$ there exists a point $$y_{x} \in K$$ farthest away from $$x$$, i.e. a point with $$\Vert y_{x} - x \Vert = \max_{y \in K} \Vert y - x\Vert.$$

My question: Is this point $$y_{x}$$ unique?

1. Second sentence:

The hyperplane $$E$$ through $$y_{x}$$ orthogonal to $$y_{x} − x$$ is then a supporting hyperplane of $$K$$ and we have $$E \cap K = \{y_{x}\}$$, hence $$y_{x} \in \operatorname{exp} K$$.

My question: Why $$E$$ supports $$K$$? I see this intuitively but I don't know how to show this. Is it a result from Support Theorem (1.4.5, page 32), because point $$y_{x}$$ is always from boundary of $$K$$? Why the intersection $$E \cap K$$ is equal to one point $$y_{x}$$? (How do we know indeed that $$y_{x}$$ is an exposed point? Is it always true?)

1. (...) On $$s$$, we can find a point $$z$$ with $$\Vert x - z \Vert > \max_{y \in \hat{K}} \Vert y - z \Vert.$$

My question: Why do we want to find that point $$z$$? Why such construction of half-line $$s$$, cube $$W$$ and ball $$B$$ is good? I don't get the idea.

1. Indeed $$y_x$$ is not unique. Consider a ball around $$x$$. The definition of $$\hat{K}$$ is therefore indeed badly formulated.

2. As $$y_x - x$$ is orthogonal to the hyperplane $$E$$, each other point on the hyperplane is further away from $$x$$ than $$y_x$$, and hence not in $$K$$.

3. We want to find a contradiction to the existence of $$x$$ in $$K \setminus \hat{K}$$. The idea is to do this by finding a point $$z$$ in $$\mathbb{R}^n$$ that is further away from $$x$$ than from any point in $$\hat{K}$$, showing that $$x$$ should be in $$\hat{K}$$ after all. A ball around $$z$$ is precisely the set of points at at most a certain distance from $$z$$. If $$\hat{K}$$ lies inside of the ball around $$z$$, and $$x$$ outside, then $$y_z \not\in \hat{K}$$, contradicting the assumption. $$s$$ and $$W$$ are just used to construct this ball.

• But how finding a point $z \in \mathbb{R}^{n}$ implies that $x$ should be in $\hat{K}$? And why point $p(\hat{K},x)$ need to be center of a facet of $W$? – apoxeiro Sep 18 '19 at 12:04
• @apoxeiro $y_z \in \hat{K}$ is a point in $K$ that lies furthest away from $z$. As $x \in K$ lies strictly further away from $z$ than $y_z$, we get a contradiction. So the assumption that there exists an $x \in K \setminus \hat{K}$ is false. It does not matter that $p(\hat{K},x)$ is the center of $W$, nor does it matter that $W$ is a cube. The crucial part is that $W$ lies inside of $B$, and $x$ lies outside of $B$. – Guus B Sep 18 '19 at 13:29

To your first question, the point is definitely not unique. Take for example $$K$$ as the unit circle and $$x$$ as the origin, then all the points in $$K$$ are at the same distance to $$x$$

• Of course, thanks! – apoxeiro Sep 17 '19 at 21:53
• It is probably unique because of convexity, though. The unit circle (without interior) is not convex. – Giuseppe Negro Sep 18 '19 at 8:48
• Yes, but if we take $K$ as closed unit ball, then for center $x$ we have infinitely many farthest points $y_{x}$ (from the boundary). So it is not unique even though $K$ is convex. Am I right? – apoxeiro Sep 18 '19 at 9:01
• This is correct! – Pebeto Sep 18 '19 at 15:19