0

Given a non-empty, closed, convex set E in Hilbert space $\mathrm{H},$ it must obtain a unique element of smallest norm.

A classic proof is use parallelogram law. But I want to show this using Zorn's lemma directly. Say let $P$ be set of all closed balls that intersects with $E$. Define ordering as containment of the balls. For each chain, an upper bound can be intersection of all the balls in this chain because intersection of all balls with $E$ is still non-empty (Show that the infinite intersection of nested non-empty closed subsets of a compact space is not empty). This gives existence of a maximal element: the smallest ball that has non-empty intersection with $E$. But this does not even uses completeness. Where went wrong? Is there any way I can do this?

Daniel Li
  • 3,200
  • 1
    Compact metric spaces are complete, so the link would use completeness. However, closed balls in a Hilbert space (but you could say normed space) of infinite dimension are never compact (except the points). –  Dec 25 '20 at 22:06
  • @Gae.S. Thank you. I get the first point. But why "balls in an infinite-dimensional Hilbert space are never compact"? The ball must be bounded and contain all limit points due to completeness assumption of Hilbert space, right? – Daniel Li Dec 25 '20 at 22:11
  • It is closed and bounded, but Riesz's lemma implies that in an infinite-dimensional normed space there is always a sequence of vectors ${v_n}_{n\in\Bbb N}$ such that $\lVert v_t\rVert<1$ for all $t$ and such that $\lVert v_n-v_m\rVert>\frac12$ for all $n\ne m$. –  Dec 25 '20 at 22:14
  • 2
    A possible method... closed convex sets in Hilbert space are weakly closed. So closed bounded convex sets are weakly compact. Can we use a finite intersection property on these? – GEdgar Dec 26 '20 at 12:04
  • @GEdgar Ah, yes, that works. Great! –  Dec 26 '20 at 12:21

1 Answers1

1

Yes, but nothing ensures that $E$ is compact. For example, let $E$ be the closed unit ball of your Hilbert space. Then it is non-empty, closed and convex, but non-compact when $H$ is infinite-dimensional. Hence, your argument with the finite-intersection property is flawed.

J. De Ro
  • 21,438