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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.

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  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.

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  • $\begingroup$ 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 $W$? $\endgroup$ – apoxeiro Sep 18 '19 at 12:04
  • $\begingroup$ @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 $B$. $\endgroup$ – Guus B Sep 18 '19 at 13:29
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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 $x$

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  • $\begingroup$ Of course, thanks! $\endgroup$ – apoxeiro Sep 17 '19 at 21:53
  • $\begingroup$ It is probably unique because of convexity, though. The unit circle (without interior) is not convex. $\endgroup$ – Giuseppe Negro Sep 18 '19 at 8:48
  • $\begingroup$ 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? $\endgroup$ – apoxeiro Sep 18 '19 at 9:01
  • $\begingroup$ This is correct! $\endgroup$ – Pebeto Sep 18 '19 at 15:19

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