1

Let $E$ be a reflexive $ \mathbb K$−Banach space and assume that $M$ is a convex, closed, bounded subset of $E.$ Prove that for any $x ∈ E$, there exists $x_0 ∈ M$ so that $|x−x_0| = \inf\{|x−y| :y ∈ M \}$.

I got the hint to use Fenchel-Rockafellar Theorem (also known as Fenchel's Duality Theorem) and then use the weak* topology $\sigma(E^*,E)$.

But I don't see the use of applying this theorem in this case. Why does one need this and how does one prove this statement using this particular theorem?

A.Γ.
  • 29,518
user372904
  • 315

1 Answers1

2

It is the classical result in convex analysis. You do not actually need Fenchel duality here if you combine the following three basic facts:

P.S. The set does not have to be bounded if you use the third mentioned variant of the Weierstrass theorem.

Community
  • 1
A.Γ.
  • 29,518
  • I find it simpler to look at $$\bigcap_{r > \operatorname{dist}(x,M)} (\overline{B_r(x)}\cap M).$$ – Daniel Fischer Nov 17 '16 at 22:49
  • @DanielFischer Yeah, it is exactly as the standard proof goes there. – A.Γ. Nov 17 '16 at 22:52
  • But I don't use the lower semicontinuity of the norm or the Weierstraß theorem, just (weak) compactness. – Daniel Fischer Nov 17 '16 at 22:54
  • @DanielFischer You suggest to prove the Weierstraß theorem instead of refering to it? Because the way it is proved is precisely looking at those sets and then take the cluster point as $x_0$. My point was to leave some details to OP to discover. – A.Γ. Nov 17 '16 at 23:01
  • Okay, thinking about it, it's the same proof as for the extreme value theorem. – Daniel Fischer Nov 17 '16 at 23:09
  • @A.G. Thanks for the hint. But from the 2nd point, how can one get that $|x-x_0|=inf{|x-y| : y \in M}$? Since they use there $liminf$ and not the infimum? And what role does the Weierstrass Theorem play here? I don't see the connection here. – user372904 Nov 18 '16 at 11:06
  • @TigerLa $\liminf$ is used in the 2d point to prove the weak lower semicontinuity of a norm by definition. Weierstrass theorem is used in the 3d point to conclude the existence of $x_0$. – A.Γ. Nov 18 '16 at 19:22
  • Okay, I will try to solve it with these 3 points. Thanks for your help! – user372904 Nov 18 '16 at 22:35
  • @TigerLa The basic idea is that 1 gives compact, 2 gives continuous and 3 gives compact+continuous=minimum exists (Weierstrass). – A.Γ. Nov 19 '16 at 06:02