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I have come across the following challenging problem in my analysis course: Let $K$ be a compact convex set in a normed linear space. Suppose that $$\sup_{x,y\in K}\{||x-y||\}=\delta>0.$$

Show there exists $x_0\in K$ such that $$\sup_{y\in K}\{||x_0-y||\}<\delta$$

We were given the following hint: since $K$ is compact, choose $a,b\in K$ such that $||a-b||=\delta$. Let $K_0$ be a maximal subset of $K$ containing $a$ and $b$ such that $||x-y||$ is either $0$ or $\delta$ whenever $x,y\in K_0$.

Now, I've made this progress so far: The set $K_0$ as described above is a discrete subset of a compact set; thus, it must be finite. My strategy so far has thus been: suppose, towards a contradiction that I can't find such an $x_0$. Then, since $K$ is compact, this means that for any $x\in K$, there is some $y\in K$ such that $||x-y||=\delta$. Using this, and convexity, I am then trying to show that I can add another point to $K_0$. I've experimented with just assuming for a bit that $K_0$ only has two elements and playing around with what this give me using the triangle inequality, but I haven't had much luck.

If anyone has some ideas on how to tackle this, it'd be much appreciated.

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1 Answer 1

$\def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|}$$K_0$ is a finite subset of $K$. As $K$ is convex, the barycentre $x_0 := \frac 1{\abs{K_0}}\sum_{x\in K_0}x$ of $K_0$ belongs to $K$. We will show that $x_0$ has the desired property. Suppose to the contrary, that there is $y \in K$ with $\norm{x_0 - y} = \delta$. We have \begin{align*} \delta &= \norm{x_0 - y}\\ &= \frac 1{\abs{K_0}}\norm{\sum_{x\in K_0} (x - y)}\\ &\le \frac 1{\abs{K_0}} \sum_{x\in K_0}\norm{x-y}\\ &\le \frac 1{\abs{K_0}} \cdot \abs{K_0}\cdot \delta\\ &= \delta. \end{align*} So we must have equality everywhere. This means, that $\norm{y - x} = \delta$ for every $x\in K_0$, especially $y \not\in K_0$. But now $K_0 \cup \{y\}$ is a proper superset of $K_0$ with the poperty that all its distances are $0$ or $\delta$, contradicting the maximality of $K_0$.

So $x_0$ is as wished.

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Thank you very much, this is fantastic. I knew in my gut that this $x_0$ must have lied in the convex hull of $K_0$. It seems so obvious now how to finish it; I guess that's just math :) –  user43645 Oct 4 '12 at 21:52

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