6
$\begingroup$

I have found the following version of Riesz-Representation theorem:

Let X be local compact hausdorff space. For any continuous functional $\psi $ on $C_0(X)$ (continuous function vanishing at infinity), there is a unique regular countably additive complex Borel measure $\mu$ on $X$ such that $$\psi(f)=\int_Xf(x)d\mu(x)$$ for all f in $C_0(X)$.

My question is how about the space of continuous bounded function class $C_b(X)$? Does there exist a similar version?

$\endgroup$
1
  • $\begingroup$ A continuous bounded function is (exactly) a function on the Stone Cech compactification, and the Riesz theorem applies to this compactification. $\endgroup$
    – Thomas
    Commented Aug 5, 2017 at 16:00

1 Answer 1

4
$\begingroup$

Not really. First, note that each functional $\psi $ on $C_b (X) $ restricts to a bounded functional on $C_0 (X) $, so that there is a (unique!) measure (regular, complex valued) with $\psi (f) =\int f d\mu $ for $ f \in C_0$.

The problem is that even though both sides of the equality make sense for $f \in C_b $, it is not necessarily true for all such $f $. As an example, consider the (closed) subspace $$ L := \{f : \Bbb {R} \to \Bbb {C} \,:\, \lim_{|x|\to \infty} f (x) \text { exists}\}, $$ and define a bounded functional on $L $ by $\psi (f) =\lim_{|x|\to\infty} f (x) $. Since $\psi $ vanishes on $C_0$, the measure from above is $\mu =0$, although $\psi (x\mapsto 1) =1 \neq 0$.

Essentially, the problem is that $C_b / C_0$ is quite large in general.

Note though that the theorem is true if $X $ is compact, simply because in this case, $C_0 = C_b $.

EDIT: Related to your additional question: One can identify the set of functionals on $C_b (X)$ with the set of regular complex measures on a suitable compactification of $X$. Indeed, if $\beta X$ denotes the Stone–Čech compactification of $X$, then each $f \in C_b (X)$ extends uniquely to a map $\tilde{f} \in C(\beta X) = C_0 (\beta X)$ (note that $f(X)$ is a bounded subset of $\Bbb{C}$, and thus contained in a compact set; in fact, this shows $\| \tilde{f} \|_\sup = \|f \|_\sup$).

Let $\iota : X \to \beta X$ be the embedding of $X$ into $\beta X$ and note that $\iota$ is a homeomorphism of $X$ onto its range in $\beta X$, since $X$ is locally compact, and thus Tychonoff. Now, if we set $$ V := \{ \tilde{f} \, : \, f \in C_b (X)\}, $$ then $\varphi : V \to \Bbb{C}, \tilde{f} \mapsto \psi(f)$ is well-defined and bounded, and by the Hahn-Banach theorem, we can extend $\varphi$ to a bounded functional on all of $C(\beta X)$. Thus, there is a regular complex measure $\nu$ on $\beta X$ satisfying $$ \psi(f)=\varphi (\tilde{f}) = \int \tilde{f} d \nu \quad \forall f \in C_b (X). $$ Note though that we integrate $\tilde{f}$, and not $f$ itself.

Conversely, each complex measure $\nu$ on $\beta X$ also induces a linear functional on $C_b (X)$ by setting $\psi(f) := \int \tilde{f} d \nu$.

In fact, $\nu$ is uniquely determined by $\psi$: It is uniquely determined once one knows $\int g d\nu$ for all $g \in C(\beta X)$. But in fact, we have $V = C(\beta X)$, since if $g \in C(\beta X)$ is arbitrary, then $f := g \circ \iota \in C_b (X)$, and $g = \tilde{f}$ is the unique extension of $g$ to a continuous map on $\beta X$.

$\endgroup$
3
  • $\begingroup$ Can I use a compactification on $X$ to achieve $C_0=C_b$? $\endgroup$ Commented Aug 5, 2017 at 17:21
  • $\begingroup$ @quallenjäger: Yes, you can, see my edit :) $\endgroup$
    – PhoemueX
    Commented Aug 5, 2017 at 19:37
  • $\begingroup$ Thank you very much! Exactly the result I was sought after. $\endgroup$ Commented Aug 5, 2017 at 19:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .