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I do not have a very strong knowledge of bipolar sets and all this stuff. Thus it could be that the question is rather easy. However I was not able to prove by myself the closedness of the following set:

We are looking at the space $L^\infty(\Omega,\mathcal{A},P)$ with a probability measure $P$. Then it is well known that the space $ba(P)$ of bounded finitely additive signed measures on $(\Omega,\mathcal{A})$, which are absolutely continuous with respect to $P$ is the dual space of $L^\infty$. Suppose $C$ is a set in $L^\infty$, which is a convex cone containing $0$ and norm closed. Therefore it is also $\sigma(L^\infty,ba)$-closed. With $C^\circ$ we denote the polar cone of $C$, which is convex and $\sigma(ba,L^\infty)$-closed. Now I'm interested in the following set:

$$K:=\{\mu\in C^\circ:\mu(\Omega)=1\}$$

This set is clearly convex, but why is it again closed in $\sigma(ba,L^\infty)$? Thanks in advance for your help.

math
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1 Answers1

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Lemma. Let $x\in X$ then the evaluation map $$ \operatorname{ev}_x:X^*\to\mathbb{C}:f\mapsto f(x) $$ is weak-$^*$ continuous.

Proof. Fix $\varepsilon>0$ and $f\in X^*$. Consider weak-$^*$ open set $U_{f,\varepsilon}=\{g\in X^*:|g(x)-f(x)|<\varepsilon\}$. Obviously for all $g\in U_{f,\varepsilon}$ we have $|\operatorname{ev}_x(g)-\operatorname{ev}_x(f)|=|g(x)-f(x)|<\varepsilon$ Hence $\operatorname{ev}_x$ is weak-$^*$ continuous.

For your particuar case evaluation map have the form $$ \operatorname{ev}_x:ba\to\mathbb{C}:\mu\mapsto\int_\Omega x(\omega)d\mu(\omega) $$ where $x\in L_\infty$. Consider set $S=\{\mu\in ba:\mu(\Omega)=1\}$. Since $\operatorname{ev}_x$ is weak-$^*$ continuous, then the set $S=(\operatorname{ev}_{\chi_\Omega})^{-1}(\{1\})$ is weak-$^*$ closed as preimage of closed set $\{1\}\subset\mathbb{C}$. Thus the set $K=C^o\cap S$ is weak-$^*$ closed as intersection of two closed sets.

Norbert
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  • Thank you for your comment. I think there is a typo: It should be $S=(\operatorname{ev}{\chi\Omega})^{-1}({1})$? Why is it obvious that the map $\operatorname{ev}_f$ is weak$^*$-continuous? – math May 17 '13 at 08:14
  • @math see edits to my answer – Norbert May 17 '13 at 12:16
  • @math: The weak*-topology on $\operatorname{ba} = (L^\infty)^\ast$ is by definition the weakest topology making all $\operatorname{ev}_f$ with $f \in L^\infty$ continuous. – Martin May 17 '13 at 15:10