Theorem on bipolars

[99]

We will find out the content of this theorem through this exercise.

Exercise: Let $E $ be a complex Hilbert space. For each nonempty subset $A\subset E $ we define its polar

$A^o=\{x\in E ~:~\forall y \in A, ~Re(x,y)\leq 1\}. $

and $(A^o)^o := A^{oo} $ is called bipolar of $A. $

1. Prove that $A^o $ is closed and convex, and $o\in A^o $ for all nonempty $A\subset E. $

2. Prove that the closed convex hull $\overline{conv(A\cup\{0\})}\subset A^{oo} $ for all nonempty $A\subset E. $

3. Let $\emptyset\neq A\subset E $ and $C = \overline{conv(A\cup \{0\})} $. Let $x\in A^{oo} $.
Prove that $Re(x-P_C(x), P_C(x)) \geq 0, $ hence $x\in C. $

Remark : $P_C(x) $ is the projection of $x $ on $C $, see [Only registered and activated users can see links. ]

4. Show that $\overline{A}=A^{oo} $ for all A convex containing $0. $

5. Let $A $ be a $\mathbb{C}- $vector subspace of $E. $ Prove that $A^{o}=A^{\perp}. $
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT] 

[99]

The most difficult item is 3.

Here is the hint :

Use the inequality in the exercise 2 of [Only registered and activated users can see links. ] for appropriate convex subsets.

[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]