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The problem:
Assume $A$ is a bounded subset of a Hilbert space $H$. Let $r$ be the infimum of the radii of closed balls containing $A$, so

$r = \inf \{s \geq 0 $ $\vert$ there exists $x \in H$ such that $\Vert y - x \Vert \leq s$ for all $y \in A \}$

  1. Show that there exists a unique closed ball of radius $r$ containing $A$. The center of this ball is called the circumcenter of $A$.

  2. Show that if $A$ is convex and closed, then the circumcenter of $A$ lies in $A$.

I don't even know where to begin with this problem. Any help is greatly appreciated!

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1 Answer 1

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For the first one, I think this might work:

Suppose that there exist two distinct balls with such property, lets say, $B(x,r)$ and $B(y,r)$. Take the balls $B(x,\delta)$ and $B(y,\delta)$ with $\delta$ small such that $$B(x,\delta)\cap B(y,\delta)\neq\emptyset$$

Now you vary $\delta$ to the first value (lets call it $s$) where $B(x,\delta)$ and $B(y,\delta)$ intersects on only one point (this is possible $H$ is a Hilbert space and hence uniformly convex). Let this point be $z$.

Can you see now that it's possible to find $r'<r$ such that the ball $B(z,r)$ cover $A$?

For the second one:

If $A$ is convex and closed the problem $$\min\{\|u-v\|,\ v\in A\}$$

is solvable for every $u\in\mathbb{R}^{n}$ and have just one solution that we gonna call $P_{A}(u)$ (projection of $u$ in $A$) and $P_{A}(u)\in A$.

Suppose that the circumcenter $u$ of $A$ is not in $A$. So the number $t=\|P_{A}(u)-u\|>0$ is well defined. This means that the ball $B(u,t)$ just intersects $A$ in $P_{A}(u)$.

On the other hand the ball $B(P_{A}(u),t)$ contains "much more" (try to give a precise meaning for this) elements of $A$, hence, you can find $r'<r$ such that the ball $B(P_{A}(u),r')$ cover $A$.

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