Functional analysis - task on convexity and dual spaces

[TaPaKaH]

Homework Statement

Let $C$ be a non-empty convex subset of a real normed space $(X,\|\cdot\|)$.
Denote $H(f,a):=\{x\in X: f(x)\leq a\}$ for $f\in X^*$ (dual space) and $a\in\mathbb{R}$.
Show that the closure $\bar{C}$ of $C$ satisfies $\bar{C}=\bigcap_{f\in X^*,a\in\mathbb{R}: C\subseteq H(f,a)}H(f,a)$.

Give an example of a real normed space $(X,\|\cdot\|)$ and a non-convex set $C$ for which the equality above does not hold.

2. Relevant information
This task comes in a homeworkset which relates to the application of the Hahn-Banach (extension) theorem, but I just can't see how one could apply it to the exercise above.

 

[micromass]

One inclusion is trivial (= doesn't require Hahn-Banach). Do you see which one that is?

Also, do you know the geometric Hahn-Banach theorem?? This is an analogous formulation of Hahn-Banach that does not deal with extensions of functionals, but rather with separation of subsets with hyperplanes. This version might be handy here.

 

[TaPaKaH]

The $\subset[\itex] inclusion seems indeed simple as the right-hand-side is a closed set (intersection of closed sets) that contains C.<br /> <br /> I know the (two-set) separation version of HBT but still can't see how that can be used in this case.$

 

[micromass]

Assume that $x\notin \overline{C}$, but assume that x is a member of the intersection. Try to find a contradiction. Start with applying the definition of the closure. This will give you an open set U disjoint from C. Does this give you ideas?

 

[TaPaKaH]

Yes, your hint makes perfect sense now.

If $x\in\bigcap_{f\in X^*,a\in\mathbb{R}: C\subseteq H(f,a)}H(f,a)$ but $x\notin\bar{C}$ then $d(x,\bar{C})=2\delta>0$ so we can separate $C$ and $U_\delta(x)$ with a hyperplane $\{f=a\}$ (for some f and a) and get a contradiction with $x\in\bigcap_{f\in X^*,a\in\mathbb{R}: C\subseteq H(f,a)}H(f,a)$.

I got the idea I was looking for, thank you very much!