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Consider the Banach space $C_b(\mathbb{R})$ of continuous bounded functions on $\mathbb{R}$ equipped with the sup-norm.

1) Do we know a precise description of its topological dual $C_b(\mathbb{R})^*$ ?

2) I was wondering what kind of relation has $C_b(\mathbb{R})^*$ with $C_c(\mathbb{R})^*$, the topological dual of the space $C_c(\mathbb{R})$ of continuous functions having compact support.

If $L\in C_b(\mathbb{R})^*$ then there exists $C_L$ such that $|L(f)|\leq C_L\|f\|_\infty$ for any $f\in C_b(\mathbb{R})$, and thus for all $f\in C_c(\mathbb{R})$, so that $L\in C_c(\mathbb{R})^*$ and $ C_b(\mathbb{R})^*\subset C_c(\mathbb{R})^*$. Is that correct ? Then, can we find all the probability measures on $\mathbb{R}$ into $C_b(\mathbb{R})^*$ ?

(In other worlds, is the "weak" topology for probability measures a weak* one ?)

2 Answers 2

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Take the algebra $\mathcal A$ generated by the closed sets in $\mathbb R$. The space of finitely-additive signed measures on $\mathcal A$, with variation norm, is the dual of $C_b(\mathbb R)$.

The top reference for this and many other interesting topics: Gillman & Jerison, Rings of Continuous Functions. Note $\mathbb R$ is metrizable, so the "zero sets" in the book are just the closed sets.

Another model for the dual: Let $\beta\mathbb R$ be the Stone-Cech compactification of $\mathbb R$, so that $C_b(\mathbb R)$ is naturally identified with $C(\beta \mathbb R)$. Then utilize your knowledge for the dual of $C(K)$ where $K$ is compact Hausdorff.

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    I understand, I was mixing "Stone-Cech" and "one-point" compactification ... Thanks again2011-11-17
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The space $C^b(\mathbb R)$ is a commutative C*-algebra, hence by the Gelfand-Naimark theorem it is *-isomorphic to some $C(K)$ space ($K$ is the spectrum of this algebra and the *-isomorphism is just the Gelfand transform). The dual of a $C(K)$ space is described by the Riesz-Kakutani theorem as the space of all regular Borel measures on $K$.

EDIT: In this case, the spectrum of $C^b(\mathbb{R})$ is just $\beta(\mathbb{R})$ which can be seen directly using basic properties of the Gelfand transform.

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    Thanks for your answer. I unfortunately don't really know such a technology ...2011-11-17