Functional analysis - task on convexity and dual spaces

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Homework Statement


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

Give an example of a real normed space [itex](X,\|\cdot\|)[/itex] and a non-convex set [itex]C[/itex] 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.
 

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  • #2
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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.
 
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The [itex]\subset[\itex] inclusion seems indeed simple as the right-hand-side is a closed set (intersection of closed sets) that contains C.

I know the (two-set) separation version of HBT but still can't see how that can be used in this case.
 
  • #4
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Assume that [itex]x\notin \overline{C}[/itex], 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?
 
  • #5
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Yes, your hint makes perfect sense now.

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

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

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