Dirac measures are dense (Hahn-Banach theorem)

Question

In an answer to this question, someone says without justification that the bidual of $C_0(\mathbb{R})$ with the weak topology is $C_0(\mathbb{R})$.

Similarly, in this question on math overflow, the answer claims that the continuous linear forms on $\mathcal{M}(I)$ are given by $C(I)$.

But, here, for example, it says that $C([0,1])$ is not reflexive.

How then are we applying the Hahn-Banach Theorem? Is there a nuance that I'm missing?

Answer 1

You have to be careful about which topology you're using. If you take ${\cal M}(I)$ with the weak-* topology as dual of $C(I)$, the continuous linear functionals on that are $C(I)$. The non-reflexivity of $C(I)$ says that there are continuous linear functionals on ${\cal M}(I)$ with the norm topology that are not given by members of $C(I)$. Such linear functionals are continuous with respect to the norm topology, but not with respect to the weak-* topology. An example of such a functional is $f(\mu) = \mu(E)$ where $E$ is a nonempty proper Borel subset of $I$.

As to the title, the linear span of Dirac measures is dense in ${\cal M}(I)$ with the weak-* topology, but not with the norm topology.