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In $\mathbb{R}^n$, a linear functional must be of the form $\sum_{i=1}^n \alpha_i x_i$ where $\alpha_i \in \mathbb{R}$.

Denote by $\mathbb{B}\left(\mathbb{R}\right)$ the set of bounded functions on $\mathbb{R}$ and by $\mathbb{CB}\left(\mathbb{R}\right)$ the set of bounded and continuous functions on $\mathbb{R}$.

Do we have the similar conclusion that a linear and continuous functional on $\mathbb{B}\left(\mathbb{R}\right)$ or $\mathbb{CB}\left(\mathbb{R}\right)$ must be of the form $\int f\left(x\right)g\left(x\right)dx$?

Ypbor
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2 Answers2

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To talk about continuity, you are not saying what topology you are giving to your spaces, which of course changes the form of the linear functionals. The canonical topology on those two spaces is that given by the supremum norm, usually denote $\|\cdot\|_\infty$.

Now, the conclusion you want cannot possibly hold as you can consider the easy example of the linear functional $$\tag1f\longmapsto f(0).$$ Still, a version of what you want does hold, in the sense that any continuous linear functional will be of the form $$ f\longmapsto \int_{\mathbb R}f\,d\mu $$ where $\mu$ is a measure (this includes the case in $(1)$, where $\mu$ is the Dirac measure). The kind of measures allowed varies slightly for both of your spaces.


When the space is $C_0(X)$ for $X$ locally compact, this is the Riesz-Markov-Kakutani Theorem. In your case you want $C_b(\mathbb R)$, and here one uses a bit of deeper theory. It turns out that $C_b(\mathbb R)$ is a unital C$^*$-algebra, and as such it can be expressed as $C(K)$ for a certain compact Hausdorff space $K$. One can show that $K=\beta\mathbb R$, the Stone-Čech Compactification of $\mathbb R$. So the dual of $C_b(\mathbb R)$ is the same as the dual of $C(\beta\mathbb R)$, which again agrees by Riesz-Markov with the space of Radon measures of bounded variation.

For the space of bounded functions, Robert has explained it very well in his answer. The dual in this case is considered a fairly nasty thing.

Martin Argerami
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In the case $B(\mathbb R)$ (with the supremum norm), the "measures" involved need only be finitely additive, but they must be defined on all subsets of $\mathbb R$, not just a $\sigma$-algebra. That is, let $\psi$ be a real-valued function on subsets of $\mathbb R$ such that

  1. $\psi(A \cup B) = \psi(A) + \psi(B)$ for any subsets $A, B$ with $A \cap B = \emptyset$.
  2. $\sup_{A \subseteq \mathbb R} |\psi(A)| < \infty$.

Corresponding to such $\psi$, we define a linear functional $\phi$ first on the subspace $S$ of simple functions, i.e. functions taking only finitely many values: if $f = \sum_{i=1}^n c_i \chi_{A_i}$ where $\chi_{A}$ is the indicator function of $A \subseteq R$, then $\phi(f) = \sum_{i=1}^n c_i \psi(A_i)$. It is easy to see that this is a linear functional on $S$ with $\|\phi\| \le \sup_{A} |\psi(A)|$. And since $S$ is dense in $B(\mathbb R)$, it extends uniquely to a continuous linear functional on $B(\mathbb R)$.

Conversely, every bounded linear functional on $B(\mathbb R)$ is of this form, i.e. for the linear functional $\phi$ we define $\psi(A) = \phi(\chi_A)$.

Finitely additive measures that are not countably additive are rather exotic beasts. See e.g. here.

Robert Israel
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