Weak convergence of continuous functions Let $X$ be an LCH space and $C_0(X)$ the set of continuous vanishing functions on $X$. If $C_0(X)$ is given the structure of a Banach space with the sup-norm, then its weak topology is given by the set of Radon measures $M(X)$ of finite total variation. One has $f_\alpha \to f$ if and only if $\int f_\alpha ~d\mu \to \int f~d\mu$ for all $\mu \in M(X)$.
My question: Is there a characterization for weak convergence in $C_0(X)$?
Weak convergence is at least as strong as pointwise convergence (because of the Dirac measures). I have an example showing that it is not the same as pointwise convergence in general.
 A: For sequences, there is the following simple characterization:

Proposition. Suppose $f_n, f \in C_0(X)$. The following are equivalent:
  
  
*
  
*$f_n \to f$ weakly;
  
*$f_n \to f$ pointwise and $\sup_n \|f_n\|_\infty < \infty$.

Proof.   $1 \implies 2$: The evaluation maps $f \mapsto f(x)$ are continuous linear functionals, so $f_n \to f$ pointwise, and $\sup_n \|f_n\| < \infty$ follows from the uniform boundedness principle.
$2 \implies 1$: If $f_n \to f$ pointwise and $\sup_n \|f_n\| < \infty$, then for any signed or complex measure $\mu$, we have $\int f_n \,d\mu \to \int f\,d\mu$ by dominated convergence, using $\sup_n \|f_n\|$ as the dominating function.  (For positive measures $\mu$, this is the classical dominated convergence theorem.  For signed measures, use the Jordan decomposition $\mu = \mu^+ - \mu^-$.  For complex measures, take real and imaginary parts.) $\quad\square$
This characterization does not work for nets.  For example, a weakly convergent net need not be bounded.
You can find this result (for $X$ compact, but the proof is the same) as Example 2 of IV.5 of Reed and Simon, volume 1 (Methods of Mathematical Physics: Functional Analysis, Revised and Enlarged Edition).
